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WELLS'   MATHEMATICAL   SERIES. 


Academic  Arithmetic. 
Academic  Algebra. 
Higher  Algebra. 
University  Algebra. 
College  Algebra- 
Plane  Geometry. 
Solid  Geometry. 
Plane  and  Solid  Geometry. 
Plane  and  Solid  Geom.etry.     Revised. 
Plane  and  Spherical  Trigonometry. 
Plane  Trigonometry. 
Essentials  of  Trigonometry. 
Logarithms  (flexible  covers). 
Elementary  Treatise  on  Logarithms. 


Special  Catalogue  and  Terms  on  application. 


THE 


ELEMENTS  OF  GEOMETRY. 


REVISED  EDITION. 


BY 


WEBSTER  WELLS,   S.B., 

Prokessou  of  Mathematics  in  the  Massachusetts 
Institute  of  Technology. 


LEACH,    SHEWELL,   &   SANBORN, 
BOSTON.  %EW  YORK.     CHICAGO. 


6^4-. 


NV  2.6 


Copyright,  1894, 
By  Webster  Wells. 


TYPOOBAPUY  UY  C.  J.  PETEB8  &  SON. 


Berwick  &  Smith,  Printees, 
Boston,  U.S.A. 


PREFACE. 


In  the  revision  of  the  author's  work  on  Plane  and 
Solid  Geometry,  many  important  improvements  have 
been  effected. 

With  a  class  just  commencing  the  study  of  Geometry, 
too  much  emphasis  cannot  be  laid  on  the  form  in  which  an 
oral  or  written  demonstration  should  he  presented. 

The  beginner  requires  a  certain  amount  of  practice  before 
he  can  acquire  the  art  of  putting  a  proof  in  a  clear  and 
logical  form. 

To  give  this  drill,  the  author  has,  through  the  whole  of 
Book  I.,  placed  directly  after  each  step  in  the  proof  the 
full  statement  of  the  reason,  in  smaller  type,  enclosed  in 
brackets. 

But  too  much  assistance  of  this  nature  is  open  to  serious 
objections,  as  it  has  a  tendency  to  make  the  pupil  a  mere 
automaton,  and  confirm  him  in  indolent  habits  of  study. 
It  has  seemed  advisable,  therefore,  in  Books  II.  to  V.,  in- 
clusive, to  give  only  the  number  of  the  section  where  the 
required  authority  is  to  be  found. 

The  above  plan  has  been  submitted  to  a  large  number  of 
representative  teachers,  and  in  nearly  every  case  has  met 
with  the  most  unqualified  approval. 

In  the  Solid  Geometry,  references  are  given  in  full  in  the 
first  sixteen  propositions  of  Book  VI.,  and  by  section  num- 
bers only  through  the  remainder  of  the  work.  On  pages 
ix,  X,  and  xi  of  the  Introduction  will  be  found  a  few  prop- 
ositions put  in  a  form  which  is  recommended  for  black- 
board work. 

iii 

184019 


iv  PREFACE. 

Particular  attention  has  been  given  to  the  arrangement 
of  the  propositions  and  corollaries  in  a  form  for  convenient 
reference.  The  statement  of  the  corollary  has  in  every  case 
been  printed  in  italics ;  and  in  nearly  every  proposition  in 
which  more  than  one  truth  is  stated,  the  various  parts  are 
distinguished  by  numerals.  Thus,  when  reference  is  made 
to  a  preceding  section,  the  pupil  will  readily  find  the  pre- 
cise statement  which  is  to  be  quoted. 

The  exercises  are  upwards  of  eight  hundred  in  number, 
and  have  been  selected  with  great  care.  In  certain  exer- 
cises which  might  otherwise  present  difficulties  to  the 
pupil,  reference  is  made  to  a  previous  section  or  exercise 
which  may  be  used  in  the  solution.  The  exercises  in  each 
Book  are  numbered  consecutively. 

In  the  Plane  Geometry,  the  new  exercises  are  largely 
numerical ;  but  in  the  Solid  Geometry,  there  is  a  consider- 
able increase  in  the  number  of  both  numerical  exercises 
and  original  theorems.  A  number  of  the  exercises  are  in 
the  nature  of  alternative  methods  of  proof  for  preceding 
propositions. 

In  the  Appendix  to  the  Plane  Geometry  will  be  found 
an  additional  set  of  exercises  of  somewhat  greater  difficulty 
than  those  previously  given. 

The  pages  have  been  arranged  in  such  a  way  as  to  avoid 
the  necessity,  while  reading  a  proof,  of  turning  the  page  for 
reference  to  the  figure. 

The  attention  of  teachers  is  specially  invited  to  the 
explanations  given  in  the  Introduction,  commencing  on 
page  vii. 

The  author  desires  to  express  his  thanks  to  the  many 
teachers,  in  all  parts  of  the  country,  who  have  furnished 
him  wi£h  valuable  suggestions  and  criticisms. 


WEBSTER  WELLS. 


Massachusetts  Institute  of  Technology, 
1894. 


CONTENTS. 


PAGE 

PllELIMINAKY   DEFINITIONS 1 


PLANE  GEOMETRY. 

BOOK      I. — Rectilinear  Figures 5 

BOOK    IL  — The  Circle 70 

BOOK  III. — Theory  OF  Proportion;  Similar  Polygons  120 

BOOK  IV.  — Areas  of  Polygons 163 

BOOK  '  V.  —  Regular  Polygons  ;  Measurement  of  tSe 

Circle 190 

APPENDIX   TO   PLANE    GEOMETRY. 

Maxima  and  Minima  of  Plane  Figures    ....  213 

Symmetrical  Figures     .        .        .        .        .        .        .        •  218 

Additional  Exercises 221 


SOLID   GEOMETRY. 
BOOK     VI.  —  Lines  and  Planes  in  Space  ;  Diedrals  ; 

POLYEDRALS 229 

BOOK    VIL  —  POLYEDRONS       .  .  .  ...  •  -204 

BOOK  VIIL— The  Cylinder,  Cone,  and  Sphere  .  .309 
BOOK     IX.  —  Measurement    of    the    Cylinder,    Cone 

AND  Sphere 348 


Answers  to  the  Numerical  Exercises. 


Of 


TO    TEACHERS. 


Among  the  most  important  objects  of  the  study  of  Geom- 
etry are  the  development  of  the  reasoning  faculties,  and  the 
cultivation  of  the  power  of  clear  and  accurate  expression. 

To  attain  these  ends,  the  pupil  should  be  required  to  state 
the  various  parts  of  a  demonstration  in  concise  and  logical 
language,  and  to  give  after  each  statement  of  a  proof  the 
reason  in  full. 

Throughout  Book  I.,  and  in  the  first  sixteen  propositions 
of  Book  Vl.,  the  required  authority  is  printed  in  each  case 
directly  after  the  statement,  in  smaller  type,  enclosed  in 
brackets. 

In  the  remaining  portions  of  the  work,  the  formal  state- 
ment of  the  reason  is  omitted,  and  there  is  given,  in  paren- 
thesis, only  the  number  of  the  section  where  the  authority 
is  to  be  found. 

In  every  such  case,  the  pupil  should  be  held,  as  in 
Book  I.,  to  the  full  statement  of  the  reason. 

The  statements  of  the  corollaries  are  in  all  cases  printed 
in  italics ;  so  that  when  a  previous  section  is  referred  to  in 
a  proof,  the  pupil  will  always  find  printed  in  italics  the 
precise  statement  to  be  quoted. 

Thus  in  Prop.  II.,  Book  II.,  reference  is  made  to  §  143; 
this  calls  for  the  following  statement :  — 

All  radii  of  a  circle  are  equal. 

While  in  general  the  complete  statement  of  the  reference 
should  be  insisted  on,  if  the  proposition  referred  to  states 
more  than  one  truth,  that  portion  only  need  be  quoted 
which  applies  to  the  case  under  consideration. 


vili  INTRODUCTION. 

Thus,  in  the  proof  of  §  29,  reference  is  made  to  §  28 ;  here 
the  complete  reference  is  not  given,  but  only  the  portion 
actually  used. 

In  most  cases,  the  various  parts  of  a  proposition  are  indi- 
cated by  numerals  ;  and  when  reference  is  made  to  a  sec- 
tion, the  numeral  following  the  number  of  the  section  shows 
which  portion  of  the  statement  is  to  be  quoted. 

Thus,  in  Prop.  XXI.,  Book  II.,  Case  I.,  reference  is  made 
to  §  83,  1 ;  this  calls  for  the  following  statement :  — 

An  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the 
two  opposite  interior  angles. 

If  a  previous  case  of  the  same  proposition  is  referred  to, 
the  reference  giyen  should  be  the  statement  of  the  theorem, 
followed  by  the  statement  of  the  previous  case. 

Thus,  on  page  92,  the  reference  "  §  189,  Case  I.'^  calls  for 
the  following  :  '■ — 

In  the  same  circle,  two  central  angles  are  in  the  same  ratio 
as  their  intercepted  arcs,  when  the  arcs  are  commensurable. 

Considerable  practice  should  be  given  in  writing  demon- 
strations on  the  blackboard ;  the  authority  for  each  state- 
ment being  given  in  full,  just  as  when  the  proof  is  given 
orally.  The  symbols  given  on  page  4  should  be  used  when- 
ever possible.  The  following  abbreviations  will  also  be 
found  of  use  :  — 


Ax., 
Def., 

Axiom. 
Definition. 

Sup., 
Alt., 

Supplementary. 
Alternate. 

Hyp., 
Cons., 

Hypothesis. 
Construction. 

Int., 
Ext., 

Interior. 
Exterior. 

Rt., 

Str., 
Adj., 

Right. 

Straight. 

Adjacent. 

Corresp. 
Rect., 

,  Corresponding. 
Rectangle. 

The  author  has  thought  that  it  would  be  an  aid  to 
teachers  to  put  a  few  propositions  in  a  form  which  is 
recommended  for  blackboard  work. 


INTRODUCTION. 


Prop.  XIII.     Book  I. 


A  straight  line  perpendicular  to  one  of  two  parallels  is  per- 
pendicular to  the  other. 


-B 


Let  the  lines  AB  and  CD  be  II  ,  and  let  ^C  be  X  to  AB. 
To  prove  AC  ±  CD. 

If  CD  is  not  ±  to  AC,  let  CE  be  drawn  A.  to  AC. 

<      .'•.  AB  II  CK 
[Two  _k  to  the  same  str,  line  are  II  .] 
But  by  hyp.,  AB  II  CD. 

.*.  CE  must  coincide  with  CD, 

[But  one  str.  line  can  be  drawn  through  a  given  point  II  to 
a  given  str.  line.] 

But  AC  1:CE,  and  .-.  AC  ±  CD. 


Prop.  XXVI.     Book  I. 

The  sum  of  the  angles  of  any  triangle  is  equal  to  tiuo  rigid 
angles. 


Let  ABC  be  any  A. 


D 


X  INTRODUCTION. 

To  prove      Z  A -{-Z  B  +Z  C  =  two  it.  A. 

Produce  AC  to  D,  and  draw  CU  ||  AB. 

Then,  Z  BCD  =  Z  A. 

[If  two  lis  are  cut  by  a  secant  line,  the  corresp.  A  are 
equal.] 

Again,  ZBCE  =  ZB. 

[If  two  lis  are  cut  by  a  secant  line,  the  alt. -int.  A  are  equal.] 

But,      Z  ECD  +  Z  BCE  -\-ZACB  =  two  rt.  A. 

[The  sum  of   all  the  A  formed  on  the  same  side  of  a 
str.  line  at  a  given  point  is  equal  to  two  rt.  A.'] 

Putting  Z  ECD  =  Z  A,  and  ZBCE  =  ZB,we  have 
Z  A  +Z  B  -^ZACB  =  two  rt.  A. 

Prop.  X.     Book  II. 

In  the.  same  circle,  or  in  equal  circles,  equal  chords  are 
equally  distant  from  the  centre. 


Let  AB  and  CD  be  equal  chords  of  the  O  ABD. 

To  prove  AB  and  CD  equally  distant  from  the  centre  0. 

Draw  OE  and  OF  ±  to  AB  and  CD,  respectively,  and 
draw  OA  and  DC. 

Then  E  is  the  middle  point  of  AB,  and  F  of  CD. 

[The  diameter  ±  to  a  chord  bisects  it.] 

Now  in  the  rt.  A  OAE  and  OCF, 
AE  =  CF, 
being  halves  of  equal  chords. 


IN.TRODUCTION. 


XI 


Also,  0A  =  oa 

[All  radii  of  a  O  are  equal.] 

.-.  AOAE  =  AOCF. 
[Two  rt.  A  are  equal  when  the  hypotenuse  and  a  leg  of 
one  are  equal  respectively  to  the  hypotenuse  and  a  leg 
of  the  other.] 

.*.  OE  =  OF. 

[In  equal  figures,  the  homologous  parts  are  equal.] 
.•.  AB  and  CD  are  equally  distant  from  0. 


Prop.  V.     Book  IV. 

The  area  of  a  triangle  is  equal   to  one-half  the  product 
of  its  base  and  altitude. 


Let  ABC  be  a  A,  having  its  altitude  AE  =  a,  and  its 
base  BC  =  h. 

To  prove  area  ABC  ~=  ^a  xh. 

Draw  AD  and  CD  II  to  BC  and  AB,  respectively,  forming 
the  O  ABCD. 

Now,  AABC=  AACD. 

[A  diagonal  of  a  O  divides  it  into  two  equal  A.] 

.'.  area  ABC  =  ^  area  ABCD. 
But,  area  ABCD  =  a  Xh. 

[The  area  of  a  O  is  equal  to  the  product  of  its  base  and 
altitude.] 

.•.  area  ABC  =  ^  a  X  b. 


GEOMETRY, 


Note.     The  attention  of  teachers  is  particularly  called  to  the 
explanations  given  in  the  Introduction,  commencing  on  p.  vii. 


PRELIMINARY  DEFINITIONS. 


71 


^ 


A  material  body. 


A  geometrical  solid. 


1.  A  material  body,  such  as  a  block  of  wood,  occupies  a 
limited  or  hounded  i)ortion  of  space. 

The  boundary  which  separates  such  a  body  from  sur- 
rounding space  is  called  the  surface  of  the  body.    - 

2.  If  the  material  composing  such  a  body  could  be  eon- 
ceived  as  taken  away  from  it,  tvlthout  altering  the  form  or 
shape  of  the  hounding  surface,  there  would  remain  d,  portion 
of  space  having  the  same  bounding  surface  as  the  former 
material  body. 

This  is  called  a  geometrical  solid,  or  simply  a  solid. 
The  bounding  surface  is  called  the  surface  of  the  solid. 

A 

3.  If  two  surfaces  cut  each  other,  their 

common  intersection  is  called  a  line. 

Thus,  if  the  surfaces  AB  and  CD  cut 
each  other,  their  common  intersection, 
EF,  is  a  line.  l^  ^'    ^i5 


GEOMETRY. 


4.  If  two  lines  cut  each  other,  their  common  intersection 
is  called  a  point.  ^  2) 

Thus,  if  the  lines  AB  and  CD  cut  each        \v^    ^-^^ 
other,  their  common  intersection,  0,  is  a  ^^^"^^^ 

point.  C^  >B 

5.  A  solid  has  extension  in  every  dh^ection  ;  but  this  is 
not  true  of  surfaces  and  lines. 

A  point  has  extension  in  no  direction,  but  simply  position 
in  space. 

6.  A  surface  may  be  conceived  as  existing  independently 
in  space,  without  reference  to  the  solid  whose  boundary  it 
forms. 

In  like  manner,  we  may  conceive  of  lines  and  points  as 
having  an  independent  existence  in  space. 

7.  A  straight  line  is  a  line  which  has  the  same  direction 
throughout  its  length  ;  as  AB. 

A  straight  line  is  also  called  a  rigid  line. 

Note.  The  word  '■'•line''''  will  be  used  hereafter  as  signifying  a 
straight  line. 

F        Q 


-B 


A  straight  line. 


A  curve. 


A  broken  line. 


A  .curved  line,  or  simply  a  curve,  is  a  line  no  portion  of 
which  is  straight ;  as  CD. 

A  broken  line  is  a  line  which  is  composed  of  different 
successive  straight  lines ;  as  EFGH. 

8.    A  plane  surface,  or  simply  a  plane,  is  a  surface  such 
that   the   straight   line    joining   any 
two  of  its  points  lies  entirely  in  the  ■^' 

surface. 

Thus,  the  surface  MN  is  a  plane        /  P 
if  the  straight  line  PQ  joining  any 
two  of  its  points  lies  entirely  in  the  surface, 


PRELIMINARY  DEFINITIONS.  3 

9.    A  curved  surface  is  a  surface  no  portion  of  which  is 
plane. 

10.  We  may  conceive  of  a  straight  line  as  being  of 
unlimited  extent  in  regard  to  length;  and  in  like  manner 
we  may  conceive  of  a  plane  as  being  of  unlimited  extent  in 
regard  to  length  and  breadth. 

11.  A  geometrical  figure  is  any  combination  of  points, 
lines,  surfaces,  or  solids. 

12.  A  plane  figure  is  a  figure  formed  by  points  and  lines 
all  lying  in  the  same  plane. 

13.  A  geometrical  figure  is  called  rectilinear,  or  right- 
lined,  when  it  is  composed  of  straight  lines  only. 

14.  Geometry  treats  of  the  properties,  construction,  and 
measurement  of  geometrical  figures. 

15.  Plane  Geometry  treats  of  plane  figures  only ;  Solid 
Geometry,  also  called  Geometry  of  Space,  or  Geometry  of 
Three  Dimensions,  treats  of  figures  which  are  not  plane. 

16.  An  Axiom  is  a  truth  assumed  as  self-evident. 

A  Theorem  is  a  truth  which  requires  demonstration. 

A  Problem  is  a  question  proposed  for  solution. 

A  Proposition  is  a  general  term  for  either  a  theorem  or  a 
problem. 

A  Postulate  is  an  assumption  of  the  possibility  of  solving 
a  certain  problem. 

A  Corollary  is  a  secondary  theorem,  which  is  an  imme- 
diate consequence  of  the  proposition  which  it  follows. 

A  Scholium  is  a  remark  or  note. 

An  Hypothesis  is  a  supposition  made  either  in  the  state- 
ment or  the  demonstration  of  a  proposition. 

One  proposition  is  said  to  be  the  Converse  of  another 
when  the  hypothesis  and  conclusion  of  the  first  are  respec- 
tively the  conclusion  and  hypothesis  of  the  second. 


4  GEOMETRY. 

17.  Postulates. 

1.  A  straight  line  can  be  drawn  between  any  two  points. 

2.  A  straight  line  can  be  produced  indefinitely  in  either 
direction. 

18.  Axioms. 

1.  Things  which  are  equal  to  the  same  thing,  or  to  equals, 
are  equal  to  each  other. 

2.  If  the  same  operation  he  performed  upjon   equals,  the 
results  will  he  equal. 

3.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

4.  The  whole  is  greater  than  any  of  its  parts. 

5.  But  one  straight  line  can  he  drawn  hetween  two  points. 

6.  A  straiglit  line  is  the  shortest  line  hetween  two  points. 

19.  A  straight  line  is  said  to  be  determined  by  any  two 
of  its  points. 

SYMBOLS. 

20.  The  following  symbols  will  be  used  in  the  work : 
X ,  multiplied  by.  Z ,  angle. 

— ,  minus.  <,  is  less  than. 

+,  plus.  >,  is  greater  than. 

=,  equals.  A,  triangle, 
o,  is  equivalent  to. 

In  addition  to  these,  the  following  are  useful  in  writing 
demonstrations  on  the  blackboard,  or  in  exercise  books  : 
A,  angles.  O,  parallelogram. 

A,  triangles.  [U,  parallelograms. 

±,  perpendicular.  0,  circle. 

Js,  perpendiculars.  (D,  circles. 

II ,  parallel.  .*.,  therefore, 

lis,  parallels. 


OP  THE    "^ 

UNIVERSITY 


PLAi^E    GEOMETRY 


Book  i. 

RECTILINEAR   FIGURES. 


DEFINITIONS  AND   GENERAL  PRINCIPLES. 

21.  If  two  straight  lines  be  drawn  from  the  same  point 
in  different  directions,  the  figure  formed 
is  called  an  Angle. 

Thus,  if  the  straight  lines  OA  and  OB 
be  drawn  from  the  same  point  0  in  dif- 
ferent dii'ections,  the  figure  AOB  is  an 
angle. 

The  point  0  is  called  the  vertex  of  the  angle,  and  the 
lines  OA  and  OB  are  called  its  sides. 

The  symbol  Z.  is  used  for  the  word  "  angle.'' 

22.  If  there  is  but  one  angle  at  a  given  vertex,  it  may  be 
designated  by  the  letter  at  that  vertex ;  but  if  two  or  more 
angles  have  the  same  vertex,  it  is  necessary,  in  order  to 
avoid  ambiguity,  to  name  also  the  letters  at  the  extremities 
of  the  sides,  placing  the  letter  at  the  vertex  between  the 
others. 

Thus,  we  should  call  the  angle  of  the  preceding  article 
"  the  angle  0 " ;  but  if  there  were  other  angles  having  the 
same  vertex,  we  should  read  it  either  AOB  or  BOA. 

Another  method  of  designating  an  angle  is  by  means  of 
a  letter  placed  between  its  sides ;  an  example  of  this  will 
be  found  in  §  71. 

5 


PLANE   GEOMETRY.  —  BOOK  I. 


23.  Two  geometrical  figures  are  said  to  be  equal  when 
one  can  be  applied  to  the  other  so  that  they  shall  coincide 
throughout. 

24.  To  prove  that  two  angles  are  equal,  it  is  not  necessary 
to  consider  the  lengths  of  their 

sides. 

Thus,  if  the  angle  ABC  can 
be  applied  to  the  angle  DEF  so 
that  the  point  B  shall  fall  on 
E,  and  the  sides  AB  and  BC  on 

J)E  and  EF^  respectively,  the  angles  are  equal,  even  if  the 
sides  AB  and  BC  are  not  equal  in  length  to  DE  and  EF, 
respectively. 


25.  Two  angles  are  called  adjacent 
when  they  have  the  same  vertex,  and  a 
common  side  between  themj  as  AOB 
and  BOC. 

26.  Two  angles  are  called  vertical,  or 
opposite,  when  the  sides  of  one  are  the 
prolongations  of  the  sides  of  the  other; 
as  DHF  and  GHE. 


PERPENDICULAR  LINES. 

27.  If  through  a  given  point  in  a  straight  line  a  line  be 
drawn  meeting  the  given  line  in  such  a  way  as  to  make  the 
adjacent  angles  equal,  each  of  the  equal  angles  is  called  a 
right  angle,  and  the  lines  are  said  to  be  perpendicular  to 
each  other. 

Thus,  if  A  be  any  point  in  the  line 
CD,  and  the  line  AB  be  drawn  in  such  a 
way  as  to  make  the  angles  BAC  and 
BAD  equal,  each  of  these  angles  is  a 

right  angle,  and  the  lines  AB  and   CD     q l jj 

are  perpendicular  to  each  other. 


RECTILINEAR  FIGURES. 


Proposition  I.     Theorem. 

28.    At  a  gwen  point  in  a  straight  line,  a  perpendicular  to 
the  line  can  he  drawn,  and  but  one. 


^ 


Let  C  be  the  given  point  in  the  straight  line  AB. 
To  prove  that  a  perpendicular  can  be  drawn  to  AB  at  C, 
and  but  one. 

Draw  CD,  making  ZBCD  <ZACJ)',  and  let  CD  be  re- 
volved about  the  point  C  as  a  pivot  towards  the  position  CA. 

Then,  Z  BCD  will  constantly  increase,  and  Z  ACD  will 
constantly  diminish ;  and  there  must  be  one  position  of  CD, 
and  only  one,  where  the  angles  are  equal. 

Let  CJEJ  be  this  position ;  then  CE  is  perpendicular  to  AB. 

Hence,  a  perpendicular  can  be  drawn  to  AB  at  C,  and 
but  one. 


0 


29.    Cor.     All  right  angles  are  eqiial. 
Let  ABC  and  DEF  be  right  angles. 
To  prove ZABC=  Z  DEF. 

Apply  Z  ABC  to  Z  DEF  in  such 
a  way  that  AB  shall  fall  upon  D^;    A 
the  point  B  falling  at  E. 

Then  BC  will   fall   upon  EF,   for  otherwise  we  should 
have  two  perpendiculars  to  DE  at  E,  which  is  impossible. 

[At  a  given  point  in  a  straight  line,  but  one  perpendicular  to  the 
line  can  be  drawn.]  (§  28.) 


B       D 


E 


Whence, 


ZABC  =  ZDEF. 


Ds 


PLANE   GEOMETRY.  —  BOOK  I. 


DEFINITIONS. 

30.  An  acute  angle  is  an  angle  which 
is  less  than  a  right  angle ;  as  ABC. 

An  obtuse  angle  is  an  angle  which  is 
greater  than  a  right  angle ;  as  DEF. 

Acute  and  obtuse  angles  "are  called 
oblique  angles ;  and  intersecting  lines 
which  are  not  j^erpendicular,  are  said  to 
be  oblique  to  each  other.  -E'  f* 

31.  An  angle  is  measured  by  finding  how  many  times  it 
contains  another  angle,  adopted  arbitrarily  as  the  unit  of 
measure. 

The  usual  unit  of  measure  is  the  degree,  which  is  the 
ninetieth  part  of  a  right  angle. 

To  express  fractional  parts  of  the  unit,  the  degree  is 
divided  into  sixty  equal  parts,  called  minutes,  and  the  min 
ute  into  sixty  equal  parts,  called  seconds. 

Degrees,  minutes,  and  seconds  are  represented  by  the 
symbols,  °,  ',  '\  respectively. 

Thus,  43°  22'  37''  represents  an  angle  of  43  degrees,  22 
minutes,  and  37  seconds. 

32.  If  the  sum  of  two. angles  is  a  right  angle,  or  90°,  one 
is  called  the  connplement  of  the  other ;  and  if  their  sum  is 
two  right  angles,  or  180°,  one  is  called  the  supplement  of 
the  other. 

Thus,  the  complement  of  an  angle  of  34°  is  90°  —  34°,  or 
56°;  and  its  supplement  is  180°  -  34°^  or  146°. 

Two  angles  which  are"  complements  of  each  other  are 
called  complementary ;  and  two  angles  which  are  supple- 
ments of  each  other  are  called  supplementary. 

33.  It  is  evident  that 

1.  The  complements  of  equal  angles  are  equal. 

2.  The  supplements  of  equal  angles  are  equal. 


I 


RECTILINEAR  FIGURES. 


EXERCISES. 


1.  How  many  degrees  are  there  in  the  complement  of  47°?  of 
83°?  of  90°? 

2.  How  many  degrees  are  there  in  the  supplement  of  31°  ?  of 
90°?  of  178°? 

Proposition  11.     Theorem. 

34.    If  one  straight  line  meet  another,  the  sicm  of  the  adja- 
cent angles  formed  is  equal  to  two  right  angles. 


Let  the  straight  line  CD  meet  the  straight  line  AB  at  C. 
To  prove   Z.  ACD -\-  A  BCD  =  two  right  angles. 

Draw  CU  perpendicular  to  AB  at  C. 

[At  a  given  point  in  a  straight  line,  a  perpendicular  to  the  line 
can  be  drawn.]  (§  28.) 

Then,   Z  ACD  +  Z  BCD  =  Z  ACE  +  Z  BCK 

But  each  of  the  angles  ACE  and  BCE  is  a  right  angle. 

Hence,  Z  ACD  -\-  Z  BCD  =  two  right  angles. 

35.  ScH.  Since  each  of  the  angles  ACD  and  BCD  is  the 
supplement  of  the  other  (§  32),  the  theorem  may  be  stated 
as  follows : 

If  one  straight   line  meet   another,  each  of  the  adjacent 
angles  formed  is  the  supplement  of  the  other. 
Such  angles  are  called  supplementary-adjacent. 

36.  Cor.  I.  The  sum  of  all  the  angles  formed  on  the  same 
side  of  a  straight  line  at  a  given  point  is  equal  to  two  right 
angles.  .  . 


10 


PLANE  GEOMETRY.  —  BOOK  I. 


37.  Cor.  II.  The  sum  of  all  the  angles  formed  about  a 
point  in  a  plane  is  equal  to  four  right  angles. 

Let  AOB,  BOC,  COD,  and  DOA  be  " 
angles  formed  about  the  point  0. 

To  prove  that  the  sum  of  the  angles 
AOB,  BOC,  COB,  and  BOA  is  equal 
to  four  right  angles. 

Produce  AO  to  E. 
Then,  the  sum  of  the  angles  AOB, 
BOC,  and  COE  is  equal  to  two  right  angles. 

[The  sum  of  all  the  angles  formed  on  the  same  side  of  a  straight 
line  at  a  given  point  is  equal  to  two  right  angles.]  (§  36.) 

In  like  manner,  the  sum  of  the  angles  EOD  and  DOA  is 
equal  to  two  right  angles. 

Therefore,  the  sum  of  the  angles  AOB,  BOC,  COD,  and 
DOA  is  equal  to  four  right  angles. 


Ex.  3.  If  in  the  above  figure  the  angles  AOB,  BOC,  and  COD 
are  respectively  49°,  88°,  and  f  of  a  right  angle,  how  many  degrees 
are  there  in  AOB  ? 

Proposition  III.     Theorem. 

38.  (Converse  of  Prop.  II.)  If  the  sum  of  two  adjacent 
angles  is  equal  to  two  right  angles,  their  exterior  sides  lie  in 
the  same  straight  line. 


E 


Let  the  sum  of  the  adjacent  angles  ACD  and  BCD  be 
equal  to  two  right  angles. 

To  prove  that  AC  and  BC  lie  in  the  same  straight  line. 


RECTILINEAR  FIGURES.  11 

• 
Let  CE  be  in  the  same  straight  line  with  AC. 
Then,  Z  ECD  is  the  supplement  of  Z  ACD. 
[If  one  straight  line  meet  another,  each  of  the  adjacent  angles 
formed  is  the  supplement  of  the  other.]  (§35.) 

But  by  hypothesis, 

Z  ^(7i>  +  Z  ^Ci)  =  two  right  angles. 
Whence,  Z  BCD  is  the  supplement  of  Z  ACD. 
Therefore,  Z  ECD  =  Z  BCD. 

[The  supplements  of  equal  angles  are  equal.]  (§  33.) 

Hence,  EC  coincides  with  BCj  and  AC  and  BC  lie  in  the 
same  straight  line. 

Proposition   IV.     Theorem. 

39.    If  two  straight  lines  intersect,  the  vertical  angles  are 
equal. 


Let  the  straight  lines  AB  and  CD  intersect  at  0. 
To  prove  AAOC^A  BOD. 

ZAOCis  the  supplement  of  Z  AOD. 

[If  one  straight  line  meet  another,  each  of  the  adjacent  angles 
formed  is  the  supplement  of  the  other.]  (§  35.) 

In  like  manner,  Z  BOD  is  the  supplement  of  Z  AOD. 
Therefore,  ZAOC  =  ZBOD. 

[The  supplements  of  equal  angles  are  equal.]  (§  33.) 

In  like  manner,  we  may  prove 

ZAOD=ZBOC. 


Ex.  4.    If  in  the  above  figure  Z  AOD  =  137°,  how  many  degrees 
are  there  in  BOC?  in  ^OC ?  in  BOD? 


12 


PLANE   GEOMETRY.  —  BOOK  I. 


Proposition   V.     Theorem. 

40.    If  a  perpendicular  he  erected  at  the  middle  point  of  a 
straight  line, 

I.    Any  poijit  in  the  perpendicular  is  eqiially  distant  from 
the  extremities  of  the  line. 

II.    Any  point  taithout  the  perpendicular  is  unequally  dis- 
tant from  the  extremities  of  the  line. 


F 

c 

A 

E. 

/  \ 

/ 

\     \ 

/ 

\^  \ 

/ 

H 

Fig.  2. 


I.  Let  CD  (Fig.  1)  be  perpendicular  to  AB  at  its  middle 
point,  D. 

Let  E  be  any  point  in  CD,  and  draw  AE  and  BE. 
To  prove  AE  =  BE. 

Let  the  figure  BDE  be  superposed  upon  ADE  by  folding 
it  over  about  DE  as  an  axis. 

We  have,  Z  BDE  =  Z  ADE. 

[All  right  angles  are  equal.]  (§29.) 

Then  the  line  BD  will  fall  upon  AD  ;  and  since,  by 
hypothesis,  BD  =  AD,  the  point  B  will  fall  at  A. 

Then  the  line  BE  will  coincide  with  AE. 

[But  one  straight  line  can  be  drawn  between  two  points.]  (Ax.  5.) 

Therefore,  AE  =  BE. 

II.  Let  CD  (Tig.  2)  be  perpendicular  to  AB  at  its  middle 
point,  D. 

Let  E  be  any  point  without  CD,  and  draw  AE  and  BF. 


d 


RECTILINEAR  FIGURES.  13 

To  prove  AF  >  BF. 

Let  AF  intersect  CD  at  E,  and  draw  BE. 

Then,  BE  +  EF  >  BF. 

[A  straight  line  is  the  shortest  line  between  two  points.]  (Ax.  6.) 

But,  BE  =  AE. 

[If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line.]  (§  40,  I.) 

Substituting  AE  for  its  equal  BE,  we  have, 

AE-\-EF>  BF, 

or,  AF>BF. 

41.  Cor.  I.  When  the  figure  BDE  is  superposed  upon 
ADE,  in-  the  proof  of  §  40,  L,  Z  EBD  coincides  with 
Z  EAD,  and  Z  BED  with  Z  AED. 

That  is,  Z  EAD  =  Z  EBD,  and  Z  AED  =  Z  ^^-D  ; 
therefore, 

If  lines  he  drawn  to  the  extremities  of  a  straight  line 
from  any  point  in  the  perpendicular  erected  at  its  middle 
point, 

1.  They  maize  equal  angles  with  the  line. 

2.  They  make  equal  angles  with  the  perjwndiciilar. 

42.  CoK.  II.  Every  point  which  is  equally  distant  from 
the  extremities  of  a  straight  line,  lies  in  the  perpendicular 
erected  at  the  middle  ptoijit  of  the  line. 

43.  CoR.  III.  A  straight  line  is  determined  by  any  two 
of  its  points  (§  19) ;  hence,  by  §  42, 

Two  points,  each  equally  distant  from, 
the  extremities  of  a  straight  line,  deter- 
Tnine  a  perpendicular  at  its  middle  point. 

Thus,  if  each  of  the  points  C  and  D  is 
equally  distant  from  A  and  B,  the  line 
CD  is  perpendicular  to  AB  at  its  middle 
point. 


14 


PLANE   GEOMETRY.  —  BOOK  I. 


Proposition  VI.     Theorem. 

44.    From  a  given  'point  without  a  straight  line,  hut  one 
perpendicular  can  be  drawn  to  the  line. 


Let  C  be  the  given  point  without  the  line  AB,  and  draw 
CD  perpendicular  to  AB. 

To  prove  that  CD  is  the  only  perpendicular  that  can  be 
drawn  from  C  to  AB. 

If  possible,  let  CE  be  another  perpendicular  from,  C 
to  AB. 

Produce  CD  to  C,  making  CD  =  CD,  and  draw  EC. 
Then  since  UD  is  perpendicular  to   CC  at  its  middle 

point  D, 

Z  CED  =  Z  C'ED. 

[If  lines  be  drawn  to  the  extremities  of  a  straight  line  from  any 
point  in  the  perpendicular  erected  at  its  middle  point,  they  make 
equal  angles  with  the  perpendicular.]  (§  41.) 

But  by  hypothesis,  Z  CED  is  a  right  angle. 

Then  its  equal,  Z  CED,  is  also  a  right  angle. 

Whence,  Z  CED  -\- Z  CED  =  two  right  angles. 

Therefore,  CEC  is  a  straight  line. 

[If  the  sum  of  two  adjacent  angles  is  equal  to  two  right  angles, 
their  exterior  sides  lie  In  the  same  straight  line.]  (§  38.) 

But  this  is  impossible,  since,  by  construction,  CDC  is  a 
straight  line. 

[But  one  straight  line  can  be  drawn  between  two  points.]  (Ax.  5.) 


Jl 


RECTILINEAR  FIGURES. 


15 


Hence,  CE  cannot  be  perpendicular  to  ABj  and  CD  is 
the  only  perpendicular  that  can  be  drawn. 

Proposition  VII.     Theorem. 

45.    The  perpendicular  is  the  shortest  line  that  can  be 
drawn  from  a  point  to  a  straight  line. 


Let  CD  be  the  perpendicular  from  C  to  the  line  AB,  and 
CE  any  other  line  drawn  from  C  to  AB. 
To  prove         *  CD  <  CE. 

Produce  CD  to  C",  making  CD  =  CD,  and  draw  EC. 
Then  since  ED  is  perpendicular  to   CC  2X  its  middle 
point  D, 

CE  =  CE. 

[If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line.]  (§  40.) 

But  CD  +  DC  <CE^  EC. 

[A  straight  line  is  the  shortest  line  between  two  points.]  (Ax.  6.) 

Therefore,  2  CD  <2  CE, 

or       .  CD  <  CE. 

46.  ScH.  The  distance  of  a  point  from  a  line  is  under- 
stood to  mean  the  length  of  the  perpendicular  from  the 
point  to  the  line. 


16  PLANE  GEOMETRY.  —  BOOK  I. 


EXERCISES. 

5.  How  many  degrees  are  there  in  the  complement,  and  in  the 
supplement,  of  an  angle  equal  to  ^  of  a  right  angle  ? 

6.  How  many  degrees  are  there  in  an  angle  whose  supplement 
is  equal  to  f  ^  of  its  complement  ? 

7.  Two  angles  are  complementary,  and  the  greater  exceeds  the 
less  by  37°.     How  many  degrees  are  there  in  each  angle  ? 

8.  Two  angles  are  supplementary,  and  the  greater  is  seven  times 
the  less.     How  many  degrees  are  there  in  each  angle  ? 

9.  Find  the  number  of  degrees  in  the  angle  the  sum  of  whose 
supplement  and  complement  is  196°. 

10.  The  straight  line  which  bisects  an  angle  bisects  also  its  verti- 
cal angle.     (§39.) 

11.  The  bisectors  of  a  pair  of  vertical  angles  lie  in  the  same 
straight  line. 

12.  The  bisectors  of  two  supplementary  adjacent  angles  are  per- 
pendicular to  each  other. 

13.  The  line  passing  through  the  vertex  of  an  angle  perpendicu- 
lar to  its  bisector  bisects  the  supplementary  adjacent  angle. 

Proposition  VIII.     Theorem; 

47.  If  two  lines  be  drawn  from  a  point  to  the  extremities 
of  a  straight  line,  their  sum  is  greater  than  the  sum  of  two 
other  lines  similarly  drawn,  but  enveloped  by  them. 


Let  the  lines  AB  and  AC  hQ  drawn  from  the  point  A  to 
the  extremities  of  the  line  BC.,  and  let  DB  and  DChQ  two 
other  lines  similarly  drawn,  but  enveloped  by  AB  and  AC. 

To  prove  AB -\- AC>  BD  +  DC. 


RECTILINEAR  FIGURES.  17 

Produce  BD  to  meet  ^  C  at  E. 

Then,  BA-\-AE>  BE. 

[A  straight  line  is  the  shortest  line  between  two  points.]  (Ax.  6.) 

Whence,  the  broken  line  BAC  is  greater  than  BEC. 

In  like  manner,      DE  +  EC  >  DC. 

Whence,  the  broken  line  BEC  is  greater  than  BDC. 

Therefore,  the  broken  line  BAC  is  greater  than  BDC. 

That  is,  AB  -{-  AC>  DB  +  DC. 

Propositiox  IX.     Theorem. 

48.    If  oblique  lines  he  drawn  from  a  point  to  a  straight 
line, 

I.  Two  oblique  lines  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular  from  the  point  to  the  line  are 
equal. 

II.    Of  tivo  oblique  lines  cutting  off  unequal  distances  from 
the  foot  of  the  perpendicular,  the  7iiore  remote  is  the  greater. 

G 


A    E  D  F    B 


I.   Let  CD  be  the  perpendicular  from  C  to  AB. 

Let  CE  and  CF  be  oblique  lines  drawn  from  C  to  AB, 
cutting  off  equal  distances  from  the  foot  of  the  per- 
pendicular. 

To  prove  CE  =  CF. 

Since  CD  is  perpendicular  to  EF  at  its  middle  point  D, 

CE  =  CF. 

[If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line.]  (§40.) 


18 


PLANE   GEOMETKY.  —  BOOK  I. 
0 


II.    Let  CD  be  the  perpendicular  from  C  to  AB. 

Let  CE  and  CF  be  oblique  lines  drawn  from  C  to  AB, 
cutting  off  unequal  distances  from  the  foot  of  the  perpen- 
dicular, CF  being  the  more  remote. 

To  prove  CF  >  CF. 

Produce  CD  to  C,  making  CD  =  CD,  and  draw  C'F 
and  CF. 

Then  since  AD  is  perpendicular  to  CC^  at  its   middle 

point  D, 

CF=  CF,  and  CF  =  CF. 

[If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line.]  (§40.) 

But,  CF  +  FC  >  CF  +  FC. 

[If  two  lines  be  drawn  from  a  point  to  the  extremities  of  a 
straight  line,  their  sum  is  greater  than  the  sum  of  two  other  lines 
similarly  drawn,  but  enveloped  by  them.]  (§  47.) 

Therefore,       2  CF  >  2  CF,  or  CF  >  CF. 

Note.  The  theorem  holds  equally  if  the  oblique  line  CE  is  on 
the  opposite  side  of  the  perpendicular  CD  from  CF. 

49.  Cor.  But  two  equal  oblique  lines  can  be  drawn  from 
a  point  to  a  straight  line. 

EXERCISES. 

14.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular,  the 
angles  are  supplementary. 

15,  A  line  drawn  through  the  vertex  of  an  angle  perpendicular 
to  its  bisector  makes  equal  angles  with  the  §ides  of  the  given  angle. 


RECTILmEAR  FIGURES.  19 

Proposition  X.     Theorem. 

50.  (Gonverse  of  Prop.  IX.,  I.)  If  oblique  lines  he  drawn 
from  a  point  to  a  straight  line,  two  equal  oblique  lines  cut  off 
equal  distances  from  the  foot  of  the  perpendicular  from  the 
point  to  the  line. 

0 


A   E 


Let  CD  be  the  perpendicular  from  C  to  AB. 
Let  CB  and  CF  be  equal  oblique  lines  drawn  from  C  to 
AB. 

To  prove  DH  =  DF. 

If  DF  were  greater  than  I>F,  CF  would  be  greater 
than  CF. 

[If  oblique  lines  be  drawn  from  a  point  to  a  straight  line,  of  two 
oblique  lines  cutting  off  unequal  distances  from  the  foot  of  the  per- 
pendicular, the  more  remote  is  the  greater.]  (§  48.) 

And  if  DF  were  less  than  DF,  CF  would  be  less  than 
CF. 

But  each  of  these  conclusions  is  contrary  to  the  hypothe- 
sis that  CF  =  CF. 

Therefore,  DF  =  DF. 

Note.  The  method  of  proof  exemplified  in  the  above  proposition 
is  known  as  the  "Indirect  Method,"  or  the  ^^  Rediictio  ad  Absur- 
dum^^ ;  the  truth  of  a  theorem  is  proved  by  supposing  it  to  be  false, 
and  showing  that  the  hypothesis  leads  to  a  false  conclusion. 

51.  Cor.  (Converse  of  Prop.  IX.,  II.)  If  two  unequal 
oblique  lines  be  drawn  from  a  point  to  a  straight  line,  the 
greater  cuts  off  the  greater  distance  from  the  foot  of  the 
perpendicular  from  the  point  to  the  line. 


20  PLANE   GEOMETRY.  —  BOOK  I. 


PARALLEL  LINES. 

52.  Definition.     Two  straight  lines  are  said  to  be  par- 
allel when  they  lie  in  the  same  plane, 

and  cannot  meet  however  far  they  may     ^ 

be  produced ;  as  AB  and  CD.  G D 

53.  Axiom.   But  one  straight  line  can  he  draivn  through 
a  given  point  parallel  to  a  given  straight  line. 

Proposition  XI.     Theorem. 

54.  Tivo  perpendiculars  to   the  same  straight 
parallel. 


Let  the  lines  AB  and  CD  be  perpendicular  to  AC. 
To  prove  AB  and  CD  parallel. 

If  AB  and  CD  are  not  parallel,  they  will  meet  in  some 
point  if  sufficiently  produced  (§  52). 

We  should  then  have  two  perpendiculars  from  the  same 
point  to  AC,  which  is  impossible. 

[From  a  given  point  without  a  straight  line,  but  one  perpendicular 
can  be  drawn  to  the  line.]  (§44.) 

Therefore,  AB  and  CD  cannot  meet,  and  are  parallel. 

Proposition   XII.     Theorem. 

55.  Two  straight  lines  parallel  to  the  same  straight  line 
are  parallel  to  each  other. 

A B 

C D 

E F 


RECTILINEAR  FIGURES.  21 

Let  the  lines  AB  and  CD  be  parallel  to  BF. 
To  prove  AB  and  CD  parallel  to  each  other. 

If  AB  and  CD  are  not  parallel,  they  will  meet  in  some 
point  if  sufficiently  produced  (§  52). 

We  should  then  have  two  lines  drawn  through  the  same 
point  parallel  to  UF,  which  is  impossible. 

[But  one  straight  line  can  be  drawn  through  a  given  point  par- 
allel to  a  given  straight  line.]  •  (§  53.) 

Therefore,  AB  and  CD  cannot  meet,  and  are  parallel. 


Proposition  XIII.     Theorem. 

56.    A  straight  line  perpendicular  to  one  of  two  parallels 
is  perpendicular  to  the  other. 


-D 


Let  the  lines  AB  and  CD  be  parallel,  and  let  ^C  be  per- 
pendicular to  AB. 

To  prove  AC  perpendicular  to  CD. 

If  CD  is  not  perpendicular  to  AC,  let  CE  be  drawn 
perpendicular  to  AC. 

Then  AB  and  CE  are  parallel. 

[Two  perpendiculars  "to  the  same  straight  line  are  parallel.]   (§  54.) 

But,  by  hypothesis,  AB  and  CD  are  parallel. 

Then  CE  must  coincide  with  CD. 

[But  one  straight  line  can  be  drawn  through  a  given  point  parallel 
to  a  given  straight  line.]  (§53.) 

But  ^C  is  perpendicular  to  CE,  and  therefore  ^C  is  per- 
pendicular to  CD. 


22 


PLANE   GEOMETRY.  —  BOOK  I. 


TRIANGLES. 

DEFINITIOlSrS. 

57.    A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines;  as  ABC.  ^ 

The  bounding  lines,  AB,  BC,  and  CA, 
are  called -the  sides  of  the  triangle,  and 
their  points  of  intersection.  A,  B,  and  C, 
are  called  the  vertices. 

The  angles  of  the  triangle  are  the 
angles  CAB,  ABC,  and  BCA,  formed  by  the  adjacent  sides. 

An  exterior  angle  of  a  triangle  is 
the  angle  formed  at  any  vertex  by 
any  side  of  the  triangle  and  the 
adjacent  side  produced;  as  ^CZ>. 

The  symbol  A  is  used  for  the 
word  "triangle." 


58.  A  triangle  is  called  scalene  when  no  two  of  its  sides 
are  equal;  isosceles  when  two  of  its  sides  are  equal;  equi- 
lateral when  all  its  sides  are  equal ;  and  equiangular  when 
all  its  angles  are  equal. 


Scalene. 


Isosceles. 


Equilateral. 


59.    A  right  triangle  is   a   triangle  which   has   a 
angle ;    as   ABC,   which    has    a    right 
angle  at  C. 

The  side  AB  opposite  to  the  right 
angle  is  called  the  hypotenuse,  and  the 
other  sides,  AC  and  BC,  the  legs. 


EECTILINEAR  FIGURES.  23 

60.  The  base  of  a  triangle  is  the  side  upon  which  it  is 
supposed  to  stand. 

In  general,  any  side  may  be  taken  as  the  base ;  but  in  an 
isosceles  triangle,  unless  otherwise  specified,  the  side  which 
is  not  one  of  the  equal  sides  is  taken  as  the  base. 

When  any  side  has  been  taken  as  the  base,  the  opposite 
angle  is  called  the  vertical  angle,  and  its 
vertex  is  called  the  vertex  of  the  triangle. 

The  altitude  of  a  triangle  is  the  per- 
pendicular drawn  from  the  vertex  to  the 
base,  produced  if  necessary. 

Thus,  in  the  triangle  ABC,  BC  is  the     ^ 
base,  BAC  the  vertical  angle,  and  AD  the  altitude. 

61.  Since  a  straight  line  is  the  shortest  line  between  two 
points  (Ax.  6),  it  follows  that 

Either  side  of  a  triangle  is  less  than  the  sum  of  the  other 
two 


Proposition  XIV.     Theorem. 

62.    Either  side  of  a  triangle  is  greater  than  the  difference 
of  the  other  two  sides. 

A 


In  the  triangle  AB C,  let  BC  be  greater  than  AC. 
To  prove  AB  >  BC  —  AC. 

We  have  AB  +  AC>  BC 

[A  straight  Une  is  the  shortest  line  between  two  points.]  (Ax.  6.) 
Subtracting  AC  from  both  members  of  the  inequality, 
AB>BC-AC. 


24  PLANE   GEOMETRY.  —  BOOK  I. 

Proposition  XV.     Theorem. 

63.    Two  triangles  are  equal  when  tivo  sides  and  the  in- 
cluded angle  of  one  are  equal  respectively  to  two  sides  a7id  the 
included  angle  of  the  other. 
G 


A  B  D  JB 

In  the  triangles  ABC  and  DEF,  let 

AB  =  DE,  AC  =  DF,  and  Z.  A  =  Z  D. 
To  prove  A  ABC  =  A  DEF. 

Superpose  the  triangle  ABC  upon  DEF  in  such  a  way 
that  Z  A  shall  coincide  with  its  equal  Z  D ;  the  side  AB 
falling  upon  DE,  and  the  side  AC  upon  DF. 

Then  since  AB  =  BE  and  AC  =  BF,  the  point  B  will 
fall  at  E,  and  the  point  C  at  F. 

Whence,  the  side  BC  will  coincide  with  EF. 

[But  one  straight  line  can  be  drawn  between  two  points.]  (Ax.  5.) 

Therefore,  ABC  and  DEF  coincide  throughout,  and  are 
equal. 

64.  Cor.  Since  ABC  and  D^i^  coincide  throughout,  we 
have 

ZB=ZE,  Z  C  =ZF,^ndLBC  =  EF. 

65.  ScH.  I.  In  equal  figures,  lines  or  angles  which  are 
similarly  placed  are  called  homologous. 

Thus,  in  the  figure  of  Prop.  XV.,  Z  A  is  homologous  to 
Z  D  ;  AB  is  homologous  to  DE ;  etc. 


66.    ScH.  II.     It  follows  from  §  65  that 

In  equal  figures,  the  homologous  pa.rts  are  equal. 


I 


RECTILINEAR  FIGURES.  25 

67.  ScH.  III.  In  equal  triangles,  the  equal  angles  lie 
opposite  the  equal  sides. 

Ex.  16.  If,  in  the  figure  of  Prop.  XV.,  AB  =  EF,  BC  =  BE,  and 
Z  B  =  Z  E,  which  angle  of  the  triangle  DEF  is  equal  to  ^  ? 
which  angle  is  equal  to  C? 

Proposition  XVI.     Theorem. 

68.  Two  triangles  are  equal  when  a  side  and  two  adjacent 
angles  of  one  are  equal  respectively  to  a  side  and  two  adja- 
cent angles  of  the  other. 

C 


In  the  triangles  ABC  and  DEF,  let 

AB  =  DE,  Z.A=ZD,2.xidi  ZB=ZE. 
To  prove  A  ABC  =  A  DEF. 

Superpose  the  triangle  ABC  upon  DEF  in  such  a  way 
that  the  side  AB  shall  coincide  with  its  equal  DE\  the 
point  A  falling  at  D,  and  the  point  B  at  E. 

Then  since  /.  A  =/.  D,  the  side  AC  will  fall  upon  DF, 
and  the  point  C  will  fall  somewhere  in  DF. 

And  since  Z  B  =  Z  E,  the  side  BC  will  fall  upon  EF, 
and  the  point  C  will  fall  somewhere  in  EF. 

Then  the  point  C,  falling  at  the  same  time  in  DF  and 
EF,  must  fall  at  their  intersection  F. 

Therefore,  ABC  and  DEF  coincide  throughout,  and  are 
equal.  

Ex.  17.   If,  in  the  figure  of  Prop.  XVI.,  AC=  BE,  Z  A=Z  F, 

and  Z  C  =  Z  D,  which  side  of  the  triangle  DEF  is  equal  to  AB  ? 
which  side  is  equal  to  BC  2 


26 


PLANE   GEOMETRY.  —  BOOK  I. 


Proposition  XVII.     Theorem. 

69.    Two  triangles  are  equal  when  the  three  sides  of  one 
are  equal  respectively  to  the  three  sides  of  the  other. 


In  the  triangles  ABC  and  DEF,  let 

AB  =  DE,  BC  =  EF,  and  CA  =  FD. 
To  prove  A  ABC  =  A  DEF. 

Place  the  triangle  DEF  in  the  position  ABF' ;  the  side 
DE  coinciding  with  its  equal  AB,  and  the  vertex  F  falling 
at  F',  on  the  opposite  side  of  AB  from  C. 

Draw  CF\ 

Then  since,  by  hypothesis,  AC  =  AF'  and  BC  =  BE', 
AB  is  perpendicular  to  CF^  at  its  middle  point. 

[Two  points,  each  equally  distant  from  the  extremities  of  a  straight 
line,  determine  the  perpendicular  at  its  middle  point.]  (§  43.) 

Whence,  Z  BAC  =  Z  BAF\ 

[If  lines  be  drawn  to  the  extremities  of  a  straight  line  from  any 
point  in  the  perpendicular  erected  at  its  middle  point,  they  make 
equal  angles  with  the  perpendicular.]  (§  41.) 

Therefore,  A  ABC  =  A  ABF\ 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other.]  (§  63.) 

That  is,  A  ABC  =  A  DEF. 


RECTILINEAR  FIGURES. 


27 


Proposition  XVIII.     Theorem. 

70.  Two  right  triangles  are  equal  when  the  hypotenuse 
and  an  adjacent  angle  of  one  are  equal  respectively  to  the 
hypotenuse  and  an  adjacent  angle  of  the  other. 

B  .E 


In  the  right  triangles  ABC  and  DUF,  let  the  hypote- 
nuse AB  be  equal  to  D£J,  and  ZA=ZI>. 

To  prove  AABC=ADI:F. 

.  Superpose  the  triangle  ABC  upon  DEF  in  such  a  way 
that  the  hypotenuse  AB  shall  coincide  with  its  equal  DE] 
the  point  A  falling  at  D,  and  the  point  B  at  E.  '     • 

Then  since  Z.  A  = /.  D,  the  side  AC  will  fall  upon  DF. 

Whence,  the  side  BC  will  fall  upon  EF. 

[From  a  given  point  without  a  straight  Une,  but  one  perpendicu- 
lar can  be  drawn  to  the  Une.]  (§44.) 

Hence,  ABC  and  DEF  coincide  throughout,  and  are  equal. 

71.  Def.  If  two  straight  lines,  AB 
and  CD,  are  cut  by  a  line  EF,  called 
a  secant  line,  the  angles  formed  are 
named  as  follows :  — 


G,  d,  e,  and  /  are  called  interior 
angles,  and  a,  h,  g,  and  h  exterior 
angles. 

G  and  /,  or  d  and  e,  are  called  alter- 
nate-interior angles. 

a  and  h,  or  h  and  g,  are  called  alternate-exterior  angles. 

a  and  e,  h  and  /,  c  and  g,  or  d  and  h,  are  called  corre- 
sponding angles. 


28  PLANE   GEOMETRY.  —  BOOK  I. 


Proposition  XIX.     Theorem. 

72.    If  two  parallels  are  cut  by  a  secant  line,  the  alternate- 
interior  angles  are  equal. 

-JB 


'H      M 


'F 


Let  the  parallels  AB  and  CD  be  cut  by  the  secant  line 
JSF  in  the  points  G  and  H,  respectively. 

To  prove  ZAGH  =  Z  GHD,  and  Zi?  6^^  =  Z  CHG. 

Through  K,  the  middle  point  of  GH,  draw  LM  perpen- 
dicular to  AB. 

Then  LM  is  also  perpendicular  to  CD. 

[A  straight  line  perpendicular  to  one  of  two  parallels  is  per- 
pendicular to  the  other.]  (§  56.) 

Then  in  the  right  triangles  GKL  and  HKM,  we  have 

GK  =  HK. 
Also,  Z  GKL  =  Z  JIKM. 

[If  two  straight  lines  intersect,  the  vertical  angles  are  equal.] 

(§39.) 
Hence,  A  GKL  =  A  HKM. 

[Two  right  triangles  are   equal  when  the  hypotenuse  and   an 

adjacent  angle  of  one  are  equal  respectively  to  the  hypotenuse  and 

*  an  adjacent  angle  of  the  other.]  (§  70.) 

Therefore,  Z  KGL  =  Z  KJI3L 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

Again,  ZAGHis  the  supplement  of  Z  BGH,  and  Z  GHD 
is  the  supplement  of  Z  CHG. 

[If  one  straight  line  meet  another,  each  of  the  adjacent  angles 
formed  is  the  supplement  of  the  other.]  (§35.) 


CTILINEAK  FIGURES.  29 


But,  ZAGH  =  Z.GHD. 

Whence,  ABGH^A  CHG. 

[The  supplements  of  equal  angles  are  equal.]  (§  33.) 

Proposition  XX.     Theorem. 

73.  (Converse  of  Prop.  XIX.)  If  two  straight  lines  are 
cut  by  a  secant  line,  malcin^  the  alternate-interior  angles 
equal,  the  two  lines  are  parallel. 


m 


■i 


Let  the  lines  AB  and  CD  be  cut  by  the  secant  line  EF 
in  the  points  G  and  H,  respectively,  making 
ZAGH  =  Z  GHD. 
To  prove  AB  and  CD  parallel. 

Through  H  draw  KL  parallel  to  AB. 
Then  since  the  parallels  AB  and  KL  are  cut  by  EF, 
ZAGH  =  Z.  GHL. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§72.) 

But  by  hypothesis,  Z.AGH=Z  GHD. 
Whence,  Z  GHL  =  Z  GHD. 

[Things  which  are  equal  to  the  same  thing  are  equal  to  each 
other.]  (Ax.  1.) 

But  this  is  impossible  unless  KL  coincides  with  CD. 

Therefore,  CD  is  parallel  to  AB. 

In  like  manner,  it  may  be  proved  that  if  AB  and  CD  are 
cut  by  EF,  making  ZBGH  =  Z  CHG,  then  AB  and  CD 
are  parallel. 


30  PLANE  GEOMETRY.  —  BOOK  I. 


Proposition  XXI.     Theorem. 

74.    If  two  parallels  are  cut  by  a  secant  linej  the  cor- 
responding angles  are  equal. 


Let  the  parallels  AB  and  CD  be  cut  by  the  secant  line 
EF  in  the  points  G  and  H,  respectively. 
To  prove  A  AGE  ^  A  CHG. 

We  have,  Z.BGH  =  A  OHG. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§'72.) 

But,  ZBGH  =  ZAGE. 

[If  two  straight  lines  intersect,  the  vertical  angles  are  equal.] 

(§39.) 
Whence,  A  AGE  =  Z.  CHG. 

[Things  which  are  equal  to  the  same  thing  are  equal  to  each 
other.]  (Ax.  1.) 

In  like  manner,  we  may  prove 

A  AGH  =  A  CHF,  A  BGE  =  A  DHG,  and 
ABGH  =  ADHF. 

75.    Cor.    If  two  'parallels  are  cut  by  a  secant  line, 

I.  The  alternate-exterior  angles  are  equal. 
For  example,  A  AGE  =  A  DHF. 

II.  The  sum  of  the  interior  angles  on  the  same  side  of  the 
secant  line  is  equal  to  two  right  angles. 

Thus,    A  AGH -{-A  CHG  =  two  right  angles. 


RECTILINEAR  FIGURES.  31 


Proposition  XXII.     Theorem. 

76.  (Converse  of  Prop.  XXI.)  If  two  straight  lines  are 
cut  by  a  secant  line,  maJcing  the  corresponding  angles  equal, 
the  two  lines  are  parallel. 


Let  the  lines  AB  and  CD  be  cut  by  the  secant  line  EF 
in  the  points  G  and  H,  respectively,  making 
ZAGE^ZCHG. 
To  prove  AB  and  CD'  parallel. 
We  have,  ZAGE  =  ZBGH. 

[If  two  straight  lines  intersect,  the  vertical  angles  are  equal.] 

(§39.) 
Whence,  ZBGH  =  ZCHG. 

Therefore,  AB  and  CD  are  parallel. 

[If  two  straight  lines  are  cut  by  a  secant  line,  making  the  alter- 
nate-interior angles  equal,  the  two  lines  are  parallel.]  (§  73.) 

In  like  manner,  it  may  be  proved  that,  if 

ZAGIf=Z  CHF,  or  Z  BGE  =  Z  DHG,  or 
ZBGH  =  ZDHF, 
then  AB  and  CD  are  parallel. 

77.    Cor.    (Converse  of  §  75.)    If  two  straight  lines  are 
cut  by  a  secant  line,  making 

I.    The  alternate-exterior  angles  equal ; 

II.    The  sum  of  the  interior  angles  on  the  same  side  of  the 
secant  line  equal  to  two  right  angles  y 
the  two  lines  are  parallel. 


32  PLANE  GEOMETRY.  —  BOOK  I. 

Proposition   XXIII.     Theorem. 
78.    Two  parallel  lines  are  everywhere  equally  distant. 
A     E  F    B 


CO  H     D 


Let  AB  and  CI>  be  parallel  lines,  and  U  and  F  any  two 
points  on  AB. 

To  prove  F  and  F  equally  distant  from  CD. 

Draw  FG  and  FJf  perpendicular  to  CD. 

Also,  draw  FG. 

Now  FG  is  perpendicular  to  AB. 

[A  straight  line  perpendicular  to  one  of  two  parallels  is  perpen- 
dicular to  the  other.]  (§  56.) 

Then  in  the  right  triangles  FFG  and  FGIT,  FG  is 
common. 

And  since  the  parallels  AB  and  CD  are  cut  by  FG, 
Z  FFG  =  Z  FGH. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§  72.) 

Hence,  A  FFG  =  A  FGH. 

[Two  right  triangles  are  equal  when  the  hypotenuse  and  an 
adjacent  angle  of  one  are  equal  respectively  to  the  hypotenuse  and 
an  adjacent  angle  of  the  other.]  (§  70.) 

Therefore,  .   FG  =  FH. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

Hence,  F  and  F  are  equally  distant  (§  46)  from  CD. 


Ex.  18.    If,  in  the  figure  of  Prop.  XIX.,  Z  ^6?^"=  68°,  how  many 
degrees  are  there  in  BGH?  in  GHB  ?  in  DHW} 


RECTILINEAR  FIGURES.  33 


Proposition   XXIV.     Theorem. 

79.  Two  angles  whose  sides  are  parallel,  each  to  each,  are 
equal  if  both  pairs  of  parallel  sides  extend  in  the  same  direc- 
tion, or  in  opposite  directions,  from  their  vertices. 


Let  AB  be  parallel  to  DH,  and  BC  to  KF. 

I.  To  prove  that  the  angles  ABC  and  DJEF,  whose  sides 
AB  and  DJS,  and  also  BC  and  FF,  extend  in  the  same 
direction  from  their  vertices,  are  equal. 

Let  BC  and  DK  intersect  at  G. 
Then  since  the  parallels  AB  and  DF  are  cut  by  BC, 
ZABC  =  ZDGC 

[If  two  parallels  are  cut  by  a  secant  line,  the  corresponding  angles 
are  equal.]  (§  74.) 

In  like  manner,     ZDGC  =  Z  DEF. 

Whence,  ZABC=ZJ)FF.  (1) 

II.  To  prove  that  the  smglesABC  and  HEK,  whose  sides 
AB  and  EH,  and  also  BC  and  EK,  extend  in  opposite 
directions  from  their  vertices,  are  equal. 

From  ( 1 ),  ZABC  =  Z  DEF. 

But,  ZDEF  =  ZHEK. 

[If  two  straight  lines  intersect,  the  vertical  angles  are  equal.] 

(§39.) 
Whence,  ZABC  =  Z  HEK. 


34  PLANE   GEOMETKY.  —  BOOK  I. 

80.    Cor.     Tivo  angles  wliose  sides  are  parallel,  each  to 
each,  are  supplementary  if  one  pair  of 
parallel  sides  extends  in  the  same  di- 
rection, and  the  other  pair  in  opposite 
directions,  from  their  vertices. 

Let  AB  be  parallel  to  DH,  and  BC 
to  KF. 

To  prove  that  the  angles  ABC  and 
DEK,  whose  sides  AB  and  DE  ex- 
tend in  the  same  direction,  and  BC  and  EK  in  opposite 
directions,  from  their  vertices,  are  supplementary. 

We  have,  /.ABC=Z  DEF. 

[Two  angles  whose  sides  are  parallel,  each  to  each,  are  equal  if 
both  pairs  of  parallel  sides  extend  in  the  same  direction  from  their 
vertices.]  (§  '79.) 

But  Z  DEF  is  the  supplement  of  Z  DEK. 
[If  one  straight  line  meet  another,  each  of  the  adjacent  angles 
formed  is  the  supplement  of  the  other.]  (§  35.) 

Whence,  ZABC  is  the  supplement  of  Z  DEK. 

EXERCISES. 

19.  If,  in  the  figure  of  Prop.  XXIV.,  Z  ABC  =  59°,  how  many 
degrees  are  there  in  each  of  the  angles  formed  about  the  point  E  ? 

20.  If  OD  and  OE  are  the  bisectors  of  two  complementary- 
adjacent  angles,  AOB  and  BOC,  how  many  degrees  are  there  in 
Z  DOEf 

21.  If  the  bisectors  of  two  adjacent  angles  make  an  angle  of  45° 
with  each  other,  the  angles  are  complementary. 

22.  If  from  a  point  0  in  a  straight  line  AB  the  lines  OC  and 
OD  be  drawn  on  opposite  sides  of  AB,  making  Z  AOC  =  ZBOD, 
prove  that  OC  and  OD  lie  in  the  same  straight  line.    (§  38.) 

23.  If,  in  a  triangle  ABC,  Z  A  =  Z  B,  a  line  parallel  to  AB 
makes  equal  angles  with  the  sides  AC  and  BC. 

24.  Two  straight  lines  are  parallel  if  any  two  points  of  either  are 
equally  distant  from  the  other. 

25.  Any  side  of  a  triangle  is  less  than  the  half-sum  of  the  sides 
of  the  triangle.     (§61.) 


RECTILINEAR  FIGURES.  35 


Proposition  XXV.     Theorem. 

81.    Two  angles  whose  sides  are  perpendicular ^  each  to 
each,  are  either  e(j[ual  or  supplementary. 


Let  AB  be  perpendicular  to  DE,  and  BC  to  FG. 

To  prove  that  Z  ABC  is  equal  to  Z  DEF,  and  supple- 
mentary to  Z  DEG. 

Draw  EH  and  EK  perpendicular  to  DE  and  EF,  respec- 
tively. 

Then  jE'^and  EKsiie  parallel  to  AB  and  BC,  respectively. 

[Two  perpendiculars  to  the  same  straight  line  are  parallel.]  (§  54.) 

Therefore,  Z  HEK  =  Z  AB  C. 

[Two  angles  whose  sides  are  parallel,  each  to  each,  are  equal  if 
both  pairs  of  parallel  sides  extend  in  the  same  direction  from  their 
vertices.]  (§  79.) 

But  each  of  the  angles  HEK  and  DEF  is  the  complement 
of  FEH. 

Whence,  Z  HEK  =  Z  DEF. 

[The  complements  of  equal  angles  are  equal.]  (§  33) 

Therefore,  ZABC=Z  DEF. 

Again,  Z  DEF  is  the  supplement  of  Z  DEG. 

[If  one  straight  line  meet  another,  each  of  the  adjacent  angles 
formed  is  the  supplement  of  the  other.]  (§35.) 

But,  ZABC  =ZDEF. 

Whence,  Z  ABC  is  the  supplement  of  Z  DEG. 

Note.  The  angles  are  equal  if  they  are  both  acute  or  both  obtuse ; 
and  supplementary  if  one  is  acute  and  the  other  obtuse. 


36  PLANE  GEOMETRY. —BOOK  I. 


Proposition  XXVI.     Theorem. 

82.    The  sum  of  the  angles  of  any  triangle  is  equal  to  two 
right  angles. 

B 


Let  ABC  be  any  triangle. 

To  prove  Z  A  -\- Z  B  -{-ZC  ==  two  right  angles. 
Produce  AC  to  D,  and  draw  CE  parallel  to  AB. 
Then  since  the  parallels  AB  and  CU  are  cut  by  AD, 
Z  ECD  =  ZA. 

[If  two  j)arallels  are  cut  by  a  secant  line,  the  corresponding  angles 
are  equal.]  (§  74.) 

Again,  ZBCE  =  ZB. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§  72.) 

But,  Z  ECD  -\-  Z  BCE  +  Z  ACB  =  two  right  angles. , 
[The  sum  of  all  the  angles  formed  on  the  same  side  of  a  straight 
line  at  a  given  point  is  equal  to  two  right  angles.]  (§  3G.) 

Putting  Z  ECD  =  Z  A,  and  ZBCE  =ZB,  we  have 
Z  A -\- Z  B  -\-ZACB  =  two  right  angles. 

83.  CoR.  I.  It  follows  from  the  above  demonstration 
that  ZBCD=ZECD-\-ZBCE  =  ZA+ZB',  hence, 

1.  An  exterior  angle  of  a  triangle  is  equal  to  the  suin  of 
the  two  opposite  inte^^ior  angles. 

2.  An  exterior  angle  of  a  triangle  is  greater  than  either  of 
the  opposite  interior  angles. 

84.  Cor.  II.  A  triangle  cannot  have  two  right  angles, 
nor  two  obtuse  angles. 


KECTILINEAR  FIGURES.  37 

85.  Cor.  III.  The  sinn  of  the  acute  angles  of  a  right 
triangle  is  equal  to  one  right  angle. 

86.  Cor.  IV.  If  tivo  triangles  have  two  angles  of  one 
equal  respectively  to  two  angles  of  the  other,  the  third  angle 
of  the  first  is  equal  to  the  third  angle  of  the  second. 

Bl.  Cor.  V.  Two  right  triangles  are  equal  when  a  leg  and 
an  acute  angle  of  one  are  equal  respectively  to  a  leg  and  the  ho- 
mologous acute  angle  of  the  other  ;  for  the  remaining  angles 
are  equal  by  §  86,  and  then  the  theorem  follows  by  §  68. 

Proposition  XXVII.     Theorem. 

88.  Two  right  triangles  are  equal  ivhen  the  hypotenuse 
and  a  leg  of  one  are  equal  respectively  to  the  hypotenuse 
and  a  leg  of  the  other. 


In  the  right  triangles  ABC  and  DEF,  let  the  hypote- 
nuse AB  be  equal  to  DE,  and  the  leg  BC  to  EF. 
To  prove  A  ABC  =  A  DEF. 

Superpose  ABC  upon  DEF  in  such  a  way  that  BC  shall 
coincide  with  EFj  the  point  B  falling  at  E,  and  C  at  F. 

Then  since  ZC  =  ZF,  AC  will  fall  upon  DF. 

But  the  equal  oblique  lines  AB  and  DE  cut  off  upon  DF 
equal  distances  from  the  foot  of  the  perpendicular  EF. 

[If  oblique  lines  be  drawn  from  a  point  to  a  straight  line,  two 
equal  oblique  lines  cut  off  equal  distances  from  the  foot  of  the  per- 
pendicular from  the  point  to  the  line.]  (§  50.) 

Whence,  the  point  A  falls  at  D. 

Hence,  ABC  and  DEF  coincide  throughout,  and  are  equal. 


38  PLANE  GEOMETKY.  — BOOK  I. 


Proposition  XXVIII.     Theorem. 

89.  If  two  triangles  have  two  sides  of  one  equal  respectively 
to  two  sides  of  the  other,  hut  the  included  angle  of  the  first 
greater  than  the  included  a^igle  of  the  second,  the  third  side 
of  the  first  is  greater  than  the  third  side  of  the  second. 


F 
In  the  triangles  ABO  and  DBF,  let 

AB  =  DE,  AC  =  DF,  smd  Z  BAC>  Z  EDF. 
To  prove  BC>  EF. 

Place  the  triangle  DEF  in  the  position  ABG\  the  side 
DE  coinciding  with  its  equal  AB,  and  the  vertex  F  falling 
at  G. 

Draw  AH  bisecting  Z  CA  G ;  also  draw  GH. 

Then  in  the  triangles  A  CH  and  A  GH,  AH  is  common. 

Also,  by  hypothesis,     AC  =  AG; 
and  by  construction,  Z  CAH  =  Z  GAH 

Therefore,  AACH=  AAGH 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other.]  (§63.) 

Whence,  CH  =  GH 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

But,  BH+GH>BG. 

[A  straight  line  is  the  shortest  line  between  two  points.]  (Ax.  6.) 

Substituting  for  GH  its  equal  CH,  we  have 

BH-\-  CH>  BG,  or  BC  >  EF. 


KECTILINEAR  FIGURES.  39 


Proposition  XXIX.     Theorem. 

90.  (Converse  of  Prop.  XXVIII.)  If  two  triangles  have 
two  sides  of  one  equal  respectively  to  two  sides  of  the  other, 
hut  the  third  side  of  the  first  greater  than  the  third  side  of 
the  second,  the  included  angle  of  the  first  is  greater  than 
the  included  angle  of  the  second. 


In  the  triangles  ABC  and  DEF,  let 

AB  =  DE,  AC  =  DF,  and  BC>  EF. 
To  prove  Z.A>  ZD. 

If  Z  A  were  equal  to  Z  D,  the  triangles  ABC  and  DEF 
would  be  equal. 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other.]  (§63.) 

Then  BC  would  be  equal  to  EF. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

Again,  \i  ZA  were  less  than  ZD,  BC  would  be  less 
than  EF. 

[If  two  triangles  have  two  sides  of  one  equal  respectively  to  two 
sides  of  the  other,  but  the  included  angle  of  the  first  greater  than 
the  included  angle  of  the  second,  the  third  side  of  the  first  is  greater 
than  the  third  side  of  the  second.]  (§  89.) 

Each  of  these  conclusions  is  contrary  to  the  hypothesis 
that  BC  i^  greater  than  EF. 

Therefore,  ZA>ZD. 


40  PLANE    GEOMETEY.  —  BOOK  I. 


Proposition  XXX.     Theorem.  ^ 

91.   In  an  isosceles  triangle,  the  angles  opposite  the  equal 
sides  are  equal. 


ABB 

Let  AC  and  BC  hQ  the  equal  sides  of  the  isosceles  tri- 
angle ABC. 

To  prove  ZA=/.B, 

Draw  CD  perpendicular  to  AB, 

Then  in  the  right  triangles  J^  (7D  and  ^(72),  CD  is  common. 
And  by  hypothesis,      AC  —  BC. 
,  Therefore,  AACD  =  ABCD. 

[Two  right  triangles  are  equal  when  the  hypotenuse  and  a  leg  of 
one  are  equal  respectively  to  the  hypotenuse  and  a  leg  of  the  other.] 

(§  88.) 
Whence,  ZA=ZB. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  QQ.) 

92.  Cor.   I.     From  the  equal  triangles  A  CD  and  BCD 
we  obtain, 

AD  =  BD,  sind  ZACD=Z  BCD. 
Hence,  the  perpendicular  from  the  vertex  to  the  base  of  an 
isosceles  triangle  bisects  the  base,  and  also  bisects  the  vertical 
angle. 

93.  Cor.  II.     An  equilateral  triangle  is  also  equiangular. 


Ex.  26.    The  angles  A  and  B  of  a  triangle  ABC  are  57°  and 
respectively;  how  many  degrees  are  there  in  the  exterior  angle  at 
the  vertex  C  ? 


i 


RECTILINEAR  FIGURES. 


41 


Proposition  XXXI.     Theorem. 

94.    (Converse  of  Prop.  XXX.)     If  two  angles  of  a  trian- 
gle are  equal,  the  sides  opj^osite  are  equal. 

C 


In  the  triangle  ABC,  let 

ZA=ZB. 

To  prove  AC  =  BC. 

Draw  CD  perpendicular  to  AB. 

Then  in  the  right  triangles  ACD  and  BCD,  CD  is  common. 

And  by  hypothesis,     ZA=ZB. 

Therefore,  A  ACD  =  A  BCD. 

[Two  right  triangles  are  equal  when  a  leg  and  an  acute  angle  of 
one  are  equal  respectively  to  a  leg  and  the  homologous  acute  angle 
of  the  other.]  (§87.) 

Whence,  AC  =  BC. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

95.  .  CoK.     An  equiangular  triangle  is  also  equilateral. 

EXERCISES. 

27.  The  angle  J5  of  a  triangle  ABC  is  three  times  the  angle  A, 
and  the  angle  C  is  five  times  the  angle  A ;  how  many  degrees  are 
there  in  each  angle  ? 

28.  The  exterior  angles  at  the  vertices  A  and  ^  of  a  triangle 
ABC  are  148°  and  83°  respectively;  how  many  degrees  are  there  in 
each  angle  of  the  triangle  ?  How  many  degrees  are  there  in  the 
exterior  angle  at  the  vertex  C  ? 

29.  How  many  degrees  are  there  in  each  angle  of  an  equiangu- 
lar triangle  ? 


42  PLANE   GEOMETRY.  —  BOOK  I. 


Proposition  XXXII.     Theosem. 

96.    In  any  triangle,  the  greater  angle  lies  opposite  the 
greater  side. 

A 


In  the  triangle  ABC,  let  ^C  be  greater  than  AB. 
To  prove  Z  ABC  >  Z  C. 

Lay  off  AD  =  AB,  and  draAv  BD. 

Then,  Z  ABD  =  Z  ABB. 

[In  an  isosceles  triangle,  the  angles  opposite  the  equal  sides  are 
equal.]  (§91.) 

But,  Z  ABB  >ZC. 

[An  exterior  angle  of  a  triangle  is  greater  than  either  of  the 
opposite  interior  angles.]  (§83.) 

Whence,  Z  ABB  >  Z  C. 

Therefore,  Z  ABC  >  Z  C. 

Proposition  XXXIII.     Theorem. 

97-  (Converse  of  Prop.  XXXII.)  In  any  triangle,  the 
greater  side  lies  opposite  the  greater  angle. 


In  the  triangle  ABC,  let  Z  ABC  be  greater  than  Z  C. 


\ 


KECTILINEAR  FIGURES.  43 

To  prove  AC>  AB. 

Draw  BD,  making  Z  CBD  =  Z  C. 
Then,  BD  =  CD. 

[If  two  angles  of  a  triangle  are  equal,  the  sides  opposite  are  equal.] 

(§94.) 
But,  AD  +  BD>  AB. 

[A  straight  line  is  the  shortest  line  between  two  points.]   (Ax.  6.) 
Substituting  for  BD  its  equal  CD,  we  have 

AD+  CD>  AB. 
That  is,  AC>AB. 

Proposition  XXXIV.     Theorem. 

98.  If  straight  lines  he  drawn  from  a  point  within  a 
triangle  to  the  extremities  of  any  side,  the  angle  included  by 
them  is  greater  than  the  angle  included  hy  the  other  two  sides. 


Let  D  be  any  point  within  the  triangle  ABC,  and  draw 
BD  and  CD. 

To  prove  Z  BDC  >ZA. 

Produce  BD  to  meet  AC  at  ^. 
Then,  ZBDC>ZDUC. 

[An  exterior  angle  of  a  triangle  is  greater  than  either  of  the 
opposite  interior  angles.]  (§  83.) 

In  like  manner,  we  have 

Z  DUC  >ZA. 
Therefore,  Z  BDC>  Z  A. 


c.  30.  Prove  Prop.  XXX.  by  drawing  CD  so  as  to  bisect  ZACB. 


44 


PLANE   GEOMETRY.  —  BOOK  I. 


Proposition   XXXV.     Theorem. 

99.    Ani/  point  in  the  bisector  of  an  angle  is  equally  dis- 
tant from  the  sides  of  the  angle. 


Let  P  be  any  point  in  the  bisector  BD  of  the  angle  ABC, 
and  draw  PM  and  PN  perpendicular  to  AB  and  BC,  res- 
pectively. 

To  prove  PM  ==  PN. 

In  the  right  triangles  BPM  and  BPN,  BP  is  common. 

And  by  hypothesis,  Z  PBM  =  Z  PBN. 

Therefore,  A  BPM  =  A  BPN. 

[Two  right  triangles  are  equal  when  the  hypotenuse  and  an  adja- 
cent angle  of  one  are  equal  respectively  to  the  hypotenuse  and  an 
udjacent  angle  of  the  other.]  (§  70.) 

Whence,  PM  =  PN. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

EXERCISES. 

31.  The  angle  at  the  vertex  of  an  isosceles  triangle  ^BC  is  equal 
to  five-thirds  the  sum  of  the  equal  angles  B  and  C.  How  many  de- 
grees are  there  in  each  angle  ? 

32.  If  the  equal  sides  of  an  isosceles  triangle  be  produced,  the 
exterior  angles  formed  with  the  base  are  equal. 

33.  The  bisector  of  the  vertical  angle  of  an  isosceles  triangle 
bisects  the  base  at  right  angles. 

34.  The  line  joining  the  vertex  of  an  isosceles  triangle  to  the 
middle  point  of  the  base,  is  perpendicular  to  the  base,  and  bisects 
the  vertical  angle. 


d 


RECTILINEAK  FIGURES.  45 


Proposition  XXXVI.    Theorem. 

100.  (Converse  of  Prop.  XXXV.)  Every  point  which  is 
within  an  a7i(/le,  and  equally  distant  from  its  sides,  lies  in 
the  bisector  of  the  angle. 


Let  the  point  P  be  within  the  angle  ABCf  and  equally 
distant  from  its  sides  ;  and  draw  BP. 
To  prove  that  BP  bisects  Z  ABC. 

DraAv  PM  and  PN  perpendicular  to  AB  and  BC,  respec- 
tively. 

Then  in  the  right  triangles  BPM  and  BPN,  BP  is  com- 
mon. 

And  by  hypothesis,     PM  =  PN. 

Therefore,  A  BPM  =  A  BPN. 

[Two  right  triangles  are  equal  when  the  hypotenuse  and  a  leg  of 
one  are  equal  respectively  to  the  hypotenuse  and' a  leg  of  the  other.] 

Whence,  Z  PBM  =  Z  PBN. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

Therefore,  BP  bisects  Z.ABC. 

EXERCISES. 

35.  The  exterior  angle  at  the  vertex  of  an  isosceles  triangle  is 
101°;  how  many  degrees  are  there  in  the  exterior  angles  at  the 
extremities  of  the  hase  ? 

36.  If  the  perpendicular  from  the  vertex  to  the  base  of  a  triangle 
bisects  the  base,  the  triangle  is  isosceles. 


46  PLANE  GEOMETRY. —BOOK  I. 

QUADRILATERALS. 
Definitions. 

101.  A  quadrilateral  is  a  portion  of  a  plane  bounded  by 
four  straight  lines ;  as  ABCD.  ^ 

The  bounding  lines  are  called  the  /\^ 

sides  of  the   quadrilateral,  and   their  /  \. 

points  of  intersection  the  vertices.  /  ^\ 

The  angles  of  the  quadrilateral  are        ^v^  / 

the   angles   formed    by  the    adjacent  ^^^^ /^ 

sides.  D 

A  diagonal  is  a  straight  line  joining  two  opposite  ver- 
tices; as  AC. 

102.  A  Trapezium  is  a  quadrilateral  no  two  of  whose 
sides  are  parallel. 

A  Trapezoid  is  a  quadrilateral  two,  and  only  two,  of 
whose  sides  are  parallel. 

A  Parallelogram  is  a  quadrilateral  whose  opposite  sides 
are  parallel. 


Trapezium.  Trapezoid.  Parallelogram. 

The  bases  of  a  trapezoid  are  its  parallel  sides ;  the  altitude 
is  the  perpendicular  distance  between  them. 

The  bases  of  a  parallelogram  are  the  side  on  which  it  is 
supposed  to  stand  and  the  parallel  side ;  the  altitude  is  the 
perpendicular  distance  between  them. 

103.  A  Rhomboid  is  a  parallelogram  whose  angles  are  not 
right  angles,  and  whose  adjacent  sides  are  unequal. 

A  Rhombus  is  a  parallelogram  whose  angles  are  not  right 
angles,  and  whose  adjacent  sides  are  equal. 

A  Rectangle  is  a  parallelogram-  whose  angles  are  right 
angles. 


I 


RECTILINEAR  FIGURES.  47 

A  Square  is  a  rectangle  whose  sides  are  equal. 


Bhomboid.  Rhombus.  Rectangle.  Square. 

Proposition  XXXVII.     Theorem. 
104.    The  opposite  sides  of  a  parallelogram  are  equal. 
B; ^0 


Let  ABCD  be  a  parallelogram. 
To  prove  AB  =  CD,  and  ^C  =  AD. 

Draw  the  diagonal  AC. 

Then  in  the  triangles  ABC  and  ACD,  AC  h  common. 
Again,  since  the  parallels  BC  and  AD  are  cut  by  AC, 
Z  BCA  =  Z  CAD. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§  72.) 

And  since  the  jmrallels  AB  and  CD  are  cut  By  AC, 
ZBAC=:ZACD. 

Therefore,  A  ABC  =  A  ACD. 

[Two  triangles  are  equal  when  a  side  and  two  adjacent  angles  of 
one  are  equal  respectively  to  a  side  and  two  adjacent  angles  of  the 
other.]  (§68.) 

Whence,  AB  =  CD,  and  BC  =  AD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

105.  Cor.  I.  Parallel  lines  included  between  parallel 
lines  are  equal. 

106.  Cor.  II.  A  diarjonal  of  a  parallelogram  divides  it 
into  two  equal  triangles. 


48  PLANE  GEOMETRY.  —  BOOK   I. 

Proposition  XXXVIII.     Theorem. 
107.    The  opposite  angles  of  a  parallelogram  are  equal. 

B . 7(7 


Let  ABCD  be  a  parallelogram. 

To  prove  Z  ^  =  Z  (7,  and  Z  7?  =  Z  D. 

Since  AB  is  parallel  to  CD,  and  AD  to  BC,  we  have 
ZA=ZC. 

[Two  angles  wliose  sides  are  parallel,  each  to  each,  are  equal  if 
both  pairs  of  parallel  sides  extend  in  ojjposite  directions  from  their 
vertices.]  (§79.) 

In  like  manner,  we  may  prove  /.  B  =  /.  D. 

Proposition  XXXIX.     Theorem. 

108.  (Converse  of  Prop.  XXXVII.)  If  the  opposite  sides 
of  a  quadrilateral  are  equal,  the  figure  is  a  parallelogram. 


In  the  quadrilateral  ABCD,  let  AB  =  CD  and  BC  =  AD. 

To  prove  ABCD  a  parallelogram. 

Draw  AC. 

Then  in  the  triangles  ABC  and  ACD,  JC  is  common. 

And  by  hypothesis,  AB  =  CD,  and  BC  =  AD. 

Therefore,        •       A  ABC  =  A  ACD. 

[Two  triangles  are  equal  when  the  three  sides  of  one  are  equal 
respectively  to  the  three  sides  of  the  other.]  (§  69.) 


RECTILINEAR  FIGURES.  49 

Whence,   ZBCA=Z  CAD,  and  Z  BAC  =  ZaCD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  Q6.) 

Then  BC  ia  parallel  to  AD,  and  AB  to  CD. 

[If  two  straight  lines  are  cut  by  a  secant  line,  making  the  alter- 
nate-interior angles  equal,  the  two  lines  are  parallel.]  (§  73.) 

Therefore,  ABCD  is  a  parallelogram. 

Proposition  XL.     Theorem. 

109.    If  two  sides  of  a  quadrilateral  are  equal  and  par- 
allel, the  figure  is  a  parallelogram. 


In  the  quadrilateral  ABCD,  let  BC  be  equal  and  parallel 
to  AD. 

To  prove  ABCD  a  parallelogram. 
Draw  AC. 

Then  in  the  triangles  ABC  and  ACD,  ^C  is  common. 
And  by  hypothesis,      BC  =  AD. 
Also,  since  the  parallels  BC  and  AD  are  cut  by  AC, 
ZBCA=Z  CAD. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§  72.) 

Therefore,  A  ABC  =  A  ACD. 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other.]  (§63.) 

Whence,  AB  =  CD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  QQ.) 

.   Therefore,  ABCD  is  a  parallelogram. 

[If  the  opposite  sides  of  a  quadrilaiteral  are  equal,  the  figure  is  a 
parallelogram.]  (§  108.) 


50  PLANE   GEOMETRY.  —  BOOK  1. 

Proposition  XLI.     Theorem. 
110.    The  diagonals  of  a  parallelogram  bisect  each  other. 

^/^ rz^c 


Let  the  diagonals  AC  and  BD  of  the  parallelogram  ABC  I) 
intersect  at  E. 

To  prove  AE  =  EC,  and  BE  =  ED. 

In  the  triangles  AED  and  BEC,  AD  =  BC. 

[The  opposite  sides  of  a  parallelogram  are  equal.]  (§  104.^1 

And  since  the  parallels  AD  and  BC  are  cut  by  AC, 

Z  EAD  =  Z  ECB. 

[If  two  jiarallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]  (§  72.) 

In  like  manner,      ZEDA=ZEBC. 

Therefore,  A  AED  =  A  BEC. 

[Two  triangles  are  equal  when  a  side  and  two  adjacent  angles  of 
one  are  equal  respectively  to  a  side  and  two  adjacent  angles  of  the 
other.]  (§68.) 

Whence,  AE  =  EC,  and  BE  =  ED. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

Note.     The  point  E  is  called  the  centre  of  the  parallelogram. 

EXERCISES. 

37.  If  one  angle  of  a  parallelogram  is  119°,  how  many  degrees 
are  there  in  each  of  the  others  ? 

38.  If  one  angle  of  a  parallelogram  is  a  right  angle,  the  figure  is 
a  rectangle. 

39.  If  from  any  point  in  the  base  of  an  isosceles  triangle  perpen- 
diculars to  the  equal  sides  be  drawn,  they  make  equal  angles  with 
the  base. 


RECTILINEAR  FIGURES.  51 


Proposition  XLII.     Theorem. 

111.    (Converse  of  Prop.  XLI.)     If  the  diagonals  of  a 
quadrilateral  bisect  each  other,  the  figure  is  a  parallelograyn. 


Let  the  diagonals  AC  and  BD  of  the  quadrilateral  ABCD 
bisect  each  other  at  E. 

To  prove  ABCD  a  .parallelogram. 

In  the  triangles  AED  and  BEC,  by  hypothesis, 

AE  =  EC,  and  BE  =  EB. 

Also,  ZAEB  =  ZBEC 

[If  two  straight  lines  intersect,  the  vertical  angles  are  equal.] 

(§39.) 
Therefore,  A  AED  =  A  BEC. 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other.]  (§63.) 

Whence,         *  AD  =  BC. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

In  like  manner,      A  AEB  =  A  CED, 
and  AB  =  CD. 

Therefore,  ABCD  is  a  parallelogram. 

[If  the  opposite  sides  of  a  quadrilateral  are  equal,  the  figure  is  a 
parallelogram.]  (§  108.) 

EXERCISES. 

40.  If  the  angles  adjacent  to  one  base  of  a  trapezoid  are  equal, 
those  adjacent  to  the  other  base  are  also  equal. 

41.  If  two  parallels  are  cut  by  a  secant  line,  the  bisectors  of  the 
four  interior  angles  form  a  rectangle. 


52  PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  XLIII.     Theorem. 

112.  Ttao  parallelograms  are  equal  when  two  adjacent 
sides  and  the  included  angle  of  one  are  equal  respectively  to 
two  adjacent  sides  and  the  included  angle  of  the  other. 


In  the  parallelograms  ABCD  and  EFGH,  let 

AB  =  EF,  AD  =  EH,  and  Z  A  =  Z  E. 
To  prove 

parallelogram  ABCD  =  parallelogram  EFGH. 

Superpose  the  parallelogram  ABCD  upon  EFGH  in  such 
a  way  that  Z  A  shall  coincide  with  its  equal  Z  E\  the  side 
AB  falling  upon  EF,  and  the  side  AD  upon  EH. 

Then  since  AB  =  EF  and  AD  =  EH,  the  point  B  will 
fall  at  F,  and  the  point  D  at  H. 

Now  since  ^C  is  parallel  to  AD,  and  EG  to  EH,  the  side 
BC  will  fall  upon  EG,  and  the  point  C  will  .fall  somewhere 
inFG. 

[But  one  straight  line  can  be  drawn  through  a  given  point  parallel 
to  a  given  straight  line.]  (§53.) 

In  like  manner,  the  side  DC  will  fall  upon  HG,  and  the 
point  C  will  fall  somewhere  in  HG. 

Therefore  the  point  C,  falling  at  the  same  time  in  EG 
and  HG,  must  fall  at  their  intersection,  G. 

Hence,  ABCD  and  EFGH  coincide  throughout,  and  are 
equal. 

113.  Cor.  Two  rectangles  are  equal  when  the  base  and 
altitude  of  one  are  equal  respectively  to  the  base  and  alti- 
tude of  the  other. 


RECTILINEAR  FIGURES. 


53 


Proposition  XLIV.     Theorem. 
114.    The  diagonals  of  a  rectangle  are  equal. 

Brc ^G 


Let  ABCD  be  a  rectangle. 

To  prove         diagonal  AC  =  diagonal  BD. 

In  the  right  triangles  ABD  and  ACD,  AD  is  common. 

Also,  AB  =CD. 

.  [The  opposite  sides  of  a  parallelogram  are  equal.]  (§  104.) 

Therefore,  A  ABD  =  A  ACD. 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other.]  (§63.) 

Whence,  AC  =  BD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

115.    Cor.     The  diagonals  of  a  square  are  equal. 

EXERCISES. 

42.  If  the  diagonals  of  a  parallelogram  are  equal,  the  figure  is  a 
rectangle. 

43.  If  two  adjacent  sides  of  a  quadrilateral  are  equal,  and  the 
diagonal  bisects  their  included  angle,  the  other  two  sides  are  equal. 

44.  The  bisectors  of  the  interior  angles  of  a  parallelogram  form  a 
rectangle. 

45.  The  bisectors  of  the  interior  angles  of  a  trapezoid  form  a 
quadrilateral,  two  of  whose  angles  are  right  angles. 

46.  If  the  angles  at  the  base  of  a  trapezoid  are  equal,  the  non- 
parallel  sides  are  also  equal! 

47.  If  the  non-parallel  sides  of  a  trapezoid  are  equal,  the  angles 
which  they  make  with  the  bases  are  equal. 


54  PLANE   GEOMETRY.  —  BOOK  I. 


Proposition  XLV.     Theorem. 

116.    The  diagonals  of  a  rhombus  bisect  each  other  at  right 
angles. 

^A ^0 


Let  ABCD  be  a  rhombus. 

To  prove  that  its  diagonals  AC  and  BD  bisect  each  other 
at  right  angles. 

Since  the  adjacent  sides  of  a  rhombus  are  equal, 

AB  =  AD,  and  BC  =  CD. 
Whence,  ^C  is  perpendicular  to  BD  at  its  middle  point. 
[Two  points,   each  equally   distant   from  the  extremities   of  a 
straight  line,  determine  a  perpendicular  at  its  middle  point.]    (§  43.) 

In  like  manner,  it  may  be  proved  that  BD  is  perpendicu- 
lar to  ^0  at  its  middle  point. 

Hence,  AC  and  BD  bisect  each  other  at  right  angles. 

POLYGONS. 
Definitions. 

117.    A  polygon  is  a  portion  of  a  plane  bounded  by  three 
or  more  straight  lines ;  as  ABCDE. 

The  bounding  lines  are  called  the 
sides  of  the  polygon,  and  their  sum  is 
called  the  perimeter. 

The  angles  of  the  polygon  are  the 
angles  EAB,  ABC,  etc.,  formed  by  the 
adjacent  sides ;  and  their  vertices  are  called  the  vertices  of 
the  polygon. 

A  diagonal  of  a  polygon  is  a  straight  line  joining  any 
two  vertices  which  are  not  consecutive;  as  AC. 


RECTILINEAR   FIGURES. 


55 


118.   Polygons  are  classified  with  reference  to  the  num- 
ber of  their  sides,  as  follows : 


No.  OF 
Sides. 

Designation. 

No.  OF 
Sides. 

Designation. 

3 
4 
5 
6 

7 

Triangle. 

Quadrilateral. 

Pentagon. 

Hexagon. 

Heptagon. 

8 

9 

10 

11 

12 

Octagon. 

Enneagon. 

Decagon. 

Undecagon. 

Dodecagon. 

119.  An  equilateral  polygon  is  a  polygon  all  of  whose 
sides  are  equal. 

An  equiangular  polygon  is  a  polygon  all  of  whose  angles 
are  equal. 

120.  A  polygon  is  called  convex  when  no  side,  if  pro- 
duced, will  enter  the  space  enclosed  by 
the  perimeter  ;  as  ABODE. 

It  is  evident  that,  in  such  a  case, 
each  angle  of  the  polygon  is  less  than 
two  right  angles. 

A  polygon  is  called  concave  when  at 
least  two  of  its  sides,  if  produced,  will 
enter  the  space  enclosed  by  the  perim- 
eter; SiS  FGHIK. 

It  is   evident  that,  in  such  a  case,  at 
least  one  angle  of  the  polygon  is  greater  than  two   right 
angles. 

Thus  in  the  polygon  FGHIK,  the  interior  angle  whose 
vertex  is  H  is  greater  than  two  right  angles. 

Such  an  angle  is  called  re-entrant. 

All  polygons  considered  hereafter  will  be  understood  to 
be  convex,  unless  the  contrary  is  stated.  ' 


56 


PLANE   GEOMETRY.  —  BOOK  I. 


121.    Two  polygons  are  said  to  be  mutually  equilateral 
when  the  sides  of  one  are 


B 


equal  respectively  to  the 
sides  of  the  other,  when 
taken  in  the  same  order. 

Thus  the  polygons  ABCD 
and  EFGH  are  mutually 
equilateral  if 

AB  =  EF,  BC  =  FG,  CD  =  GH,  and  DA  =  HE. 


122.  Two  polygons  are  said  to  be  mutually  equiangular 
when   the  angles  of  one   are 

equal  respectively  to  the  an- 
gles of  the  other,  when  taken 
in  the  same  order. 

Thus  the  polygons  ABCD 
and  EFGH Siie  mutually  equi- 
angular if 

ZA=ZE,  ZB=ZF,  ZC  =  ZG,£indZD=ZH. 

123.  In  polygons  which  are  mutually  equilateral  or 
mutually  equiangular,  sides  or  angles  which  are  similarly 
placed  are  called  homologous. 

If  two  triangles  are  mutually  equilateral,  they  are  also 
mutually  equiangular  (§  69) ;  but  with  this  exception,  two 
polygons  may  be  mutually  equilateral  without  being  mutu- 
ally equiangular,  or  mutually  equiangular  without  being 
mutually  equilateral. 

124.  If  two  polygons  are  both  mutually  equilateral  and 
mutually  equiangular,  tliey  are  equal. 

For  they  can  evidently  be  applied  one  to  the  other  so  as 
to  coincide  throughout. 

125.  Two  polygons  are  equal  when  they  are  composed  of 
the  same  number  of  triangles,  equal  each  to  each,  and  simi- 
larly placed;  for  they  can  evidently  be  applied  one  to  the 
other  so  as  to  coincide  throughout. 


RECTILINEAR  FIGURES.  57 

Proposition  XLVI.     Theorem. 

126.  The  sum  of  the  angles  of  any  polygon  is  equal  to 
two  rig  Jit  angles  taken  as  many  times,  less  two,  as  the  poly- 
gon has  sides. 


Any  polygon  may  be  divided  into  triangles  by  drawing 
diagonals  from  one  of  its  vertices ;  the  number  of  triangles 
being  equal  to  the  number  of  sides  of  the  polygon,  less 
two. 

The  sum  of  the  angles  of  the  polygon  is  equal  to  the  sum 
of  the  angles  of  the  triangles. 

And  the  sum  of  the  angles  of  each  triangle  is  equal  to 
two  right  angles. 

[The  sum  of  the  angles  of  any  triangle  is  equal  to  two  right 
angles.]  (§  82.) 

Hence,  the  sum  of  the  angles  of  the  polygon  is  equal  to 
two  right  angles  taken  as  many  times,  less  two,  as  the 
polygon  has  sides. 

127.  Cor.  I.  The  sum  of  the  angles  of  any  polygon  is 
equal  to  tivice  as  many  right  angles  as  the  polygon  has 
sides,  less  four  right  angles. 

Thus,  if  B  represents  a  right  angle,  and  n  the  number  of 
sides  of  a  polygon,  the  sum  of  its  angles  is  expressed  by 
2nR-4:E. 

128.  CoR.  II.  The  sum  of  the  angles  of  a  quadrilat- 
eral is  equal  to  four  right  angles;  of  a  pentagon,  six  right 
angles  ;  of  a  hexagon,  eight  right  angles  /  etc. 


58  PLANE   GEOMETRY.  —  BOOK  I. 


PuoposiTiON  XLVII.     Theorem. 


\ 


129.  If  the  sides  of  any  polygon  be  produced  so  as  to  form 
an  exterior  angle  at  each  vertex,  the  sum  of  these  exterior 
angles  is  equal  to  four  right  angles. 


The  sum  of  the  exterior  and  interior  angles  formed  at 
any  one  vertex  is  equal  to  two  right  angles. 

[If  one  straight  line  meet  another,  the  sum  of  the  adjacent  angles 
formed  is  equal  to  two  right  angles.]  (§  34.) 

Hence,  the  sum  of  all  the  exterior  and  interior  angles  is 
equal  to  twice  as  many  right  angles  as  the  polygon  has 
sides. 

But  the  sum  of  tlie  interior  angles  alone  is  equal  to  twice 
as  many  right  angles  as  the  polygon  has  sides,  less  four 
right  angles. 

[The  sum  of  the  angles  of  any  polygon  is  equal  to  twice  as  many 
right  angles  as  the  polygon  has  sides,  less  four  right  angles.]  (§  127.) 

Whence,  the  sum  of  the  exterior  angles  is  equal  to  four 
right  angles. 

EXERCISES. 

48.  How  many  degrees  are  there  in  each  angle  of  an  equiangular 
hexagon  ?  of  an  equiangular  octagon  ?  of  an  equiangular  decagon  ? 
of  an  equiangular  dodecagon  ? 

49.  How  many  degrees  are  there  in  the  exterior  angle  at  each 
vertex  of  an  equiangular  pentagon  ? 

50.  If  two  angles  of  a  quadrilateral  are  supplementary,  the  other 
two  angles  are  supplementary. 


RECTILINEAR  FIGURES.  69 

MISCELLANEOUS  THEOREMS. 

Proposition  XLVIII.     Theorem. 

130.  The  line  joining  the  middle  points  of  two  sides  of  a 
trianf/le  is  parallel  to  the  third  side,  and  equal  to  one-half 
of  it. 

A 


Let  DU  bisect  the  sides  AB  and  ^C  of  the  triangle  ABC. 
To  prove  BU  parallel  to  BC,  and  equal  to  ^BC. 

Draw  BF  parallel  to  AC,  meeting  JSI)  produced  at  F. 
Then  in  the  triangles  ADF  and  BBF,  ZA  =Z  DBF. 

[If  two  parallels  are  cut  by  a  secant  line,  the  alternate-interior 
angles  are  equal.]'  (§72.) 

Also,  Z  ABF  =  Z  BDF. 

[If  two  straight  lines  intersect,  the  vertical  angles  are  equal.] 

(§  39.) 
And  by  hypothesis,      AD  =  BD. 

Therefore,  A  ADE  =  A  BDF. 

[Two  triangles  are  equal  when  a  side  and  two  adjacent  angles 
of  one  are  equal  respectively  to  a  side  and  two  adjacent  angles  of 
the  other.]  (§  68.) 

Whence,  DE  =  DF,  and  AE  =  BF. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  06.) 

Then,  since  AE  =  EC,  BF  is  equal  and  parallel  to  CE. 
Whence,  BCEF  is  a  parallelogram. 

[If  two  sides  of  a  quadrilateral  are  equal  and  parallel,  the  figure 
is  a  parallelogram.]  (§  109.) 

Hence,  DE  is  parallel  to  BC,  an(i  DE  =  ^  FE  =  i  BC. 
[The  opposite  sides  of  a  parallelogram  are  equal.]  (§  104.) 


60 


PLANE   GEOMETRY.  —  BOOK  I. 


^  131.    Cor.    The  line  which  bisects  one  side  of  a 
and  is  parallel  to  another  side,  bisects 
also  the  third  side. 

In  the  triangle  ABC,  let  DE  be  par- 
allel to  BC,  and  bisect  AB. 

To  prove  that  DE  bisects  AC. 

A  line  drawn  from  D  to  the  middle 
point  of  AC  will  be  parallel  to  BC. 

[The  line  joining  the  middle  points  of  two  sides  of  a  triangle 
is  parallel  to  the  third  side.]  (§  130.) 

Then  this  line  will  coincide  with  DE. 

[But  one  straight  line  can  be  drawn  through  a  given  point  parallel 
to  a  given  straight  line.]  (§53.) 

Therefore,  DE  bisects  AC. 


EXERCISES. 

51.  The  bisectors  of  the  equal  angles  of  an  isosceles  triangle  form, 
with  the  base,  another  isosceles  triangle. 

52.  Either  exterior  angle  at  the  base  of  an  isosceles  triangle  is 
equal  to  the  sum  of  a  right  angle  and  one-half  the  vertical  angle. 

53.  The  straight  lines  bisecting  the  equal  angles  of  an  isosceles 
triangle,  and  terminating  in  the  opposite  sides,  are  equal. 

54.  The  perpendiculars  from  the  extremities  of  the  base  of  an 
isosceles  triangle  to  the  opposite  sides  are  equal. 

55.  The  angle  between  the  bisectors  of  the  equal  angles  of  an 
isosceles  triangle  is  equal  to  the  exterior  angle  at  the  base  of  the 
triangle. 

56.  If  through  a  point  midway  between  two  parallels  two  secant 
lines  be  drawn,  they  intercept  equal  portions  of  the  parallels, 

57.  If  a  line  joining  two  parallels  be  bisected,  any  line  drawn 
through  the  point  of  bisection  and  included  between  the  parallels 
will  be  bisected  at  the  point. 

58.  If  perpendiculars  BE  and  DF  be  drawn  from  the  vertices  B 
and  D  of  the  parallelogram  ABCI)  to  the  diagonal  AC,  prove  that 
BE  =  DF. 

59.  The  lines  joining  the  middle  points  of  the  sides  of  a  triangle 
divide  it  into  four  equal  triangles.     (§  130.) 


RECTILINEAR  FIGURES.  61 


Proposition   XLIX.    Theorem. 

132.  The  line  joining  the  middle  points  of  the  non-paral- 
lel sides  of  a  trapezoid  is  parallel  to  the  bases,  and  equal  to 
one-half  their  sum. 


Let  the  line  EF  bisect  the  non-parallel  sides  AB  and  CD 
of  the  trapezoid  ABCD. 

I.  To  prove  EP  parallel  to  AD  and  BC. 

Let  EK  be  parallel  to  AD  and  BC;  and  draw  BD,  inter- 
secting EF  at  G,  and  EK  at  H. 

Then  in  the  triangle  ABD,  EH  is  parallel  to  AD  and 
bisects  AB ;  it  therefore  bisects  BD. 

[The  line  which  bisects  one  side  of  a  triangle,  and  is  parallel  to 
another  side,  bisects  also  the  third  side.]  (§  131.) 

In  like  manner,  in  the  triangle  BCD,  UK  is  parallel  to 
BC  and  bisects  BD;  it  therefore  bisects  CD. 

But  this  is  impossible  unless  EK  coincides  with  EF. 
Hence,  EF  is  parallel  to  AD  and  BC. 

II.  To  prove  EF  =  ^  (AD+BC). 
Since  EG  bisects  AB  and  BD, 

EG=  \AD.  (1) 

[The  line  joining  the  middle  points  of  two  sides  of  a  triangle 

is  equal  to  one-half  the  third  side.]  (§  130.) 
And  since  GF  bisects  BD  and  CD, 

GF  =  ^BC.  (2) 

Adding  (1)  and  (2),  we  have, 

EG-\-  GF  =  \AD  +  ^BC. 
That  is,  EF  =  ^  {AD  +  BC). 

OF  THE     "*^ 


62  PLANE   GEOMETEY.  — BOOK  I. 

133.  Cor.  The  line  which  is  parallel  to  the  bases  of  a 
trapezoid,  and  bisects  one  of  the  non-pjarallel  sides,  bisects  the 
other  also. 

Proposition  L.     Theorem. 

134.  The  bisectors  of  the  angles  of  a  triangle  meet  in  a 
common  point. 


Let  AD,  BU,SiiidL  CF  be  the  bisectors  of  the  angles  Ay 
B,  and  C  of  the  triangle  ABC. 

To  prove  that  AD,  BE,  and  CF  intersect  in  a  common 
point. 

Let  AD  and  BE  intersect  at  0. 

Then  since  0  is  in  the  bisector  AD,  it  is  equally  distant 
from  the  sides  AB  and  AC. 

[Any  point  in  the  bisector  of  an  angle  is  equally  distant  from  the 
sides  of  the  angle.]  (§  99.) 

In  like  manner,  since  0  is  in  the  bisector  BE,  it  is 
equally  distant  from  the  sides  AB  and  BC. 

Then  0  is  equally  distant  from  the  sides  AC  and  BC, 
and  therefore  lies  in  the  bisector  CF. 

[Every  point  which  is  within  an  angle,  and  equally  distant  from 
its  sides,  lies  in  the  bisector  of  the  angle.]  (§  100.) 

Hence,  AD,  BE,  and  CF  meet  in  the  common  point  0. 

135.  Cor.  The  point  of  intersection  of  the  bisectors  of 
the  angles  of  a  t7'iangle  is  equally  distant  from  the  sides 
of  the  triangle. 


RECTILINEAR  FIGURES.  63 


Proposition  LI.     Theorem. 


136.    The  perpendiculars  erected  at  the  middle  points  of 
the  sides  of  a  triangle  meet  in  a  common  point. 


Let  DG,  EH,  and  FK  be  the  perpendiculars  erected  at 
the  middle  points  of  the  sides  BC,  CA,  and  AB,  of  the 
triangle  ABC. 

To  prove  thsit  DG,  EH,  and  FK  intersect  in  a  common 
point. 

Let  DG  and  EH  intersect  at  0. 

Then  since  0  is  in  the  perpendicular  DG,  it  is  equally 
distant  from  B  and  C. 

[If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the  ex- 
tremities of  the  line.]  (§  40.) 

In  like  manner,  since  0  is  in  the  perpendicular  EH,  it  is 
equally  distant  from  A  and  C. 

Then  0  is  equally  distant  from  A  and  B,  and  therefore 
lies  in  the  perpendicular  FK. 

[Every  point  which  is  equally  distant  from  the  extremities  of  a 
straight  line,  lies  in  the  perpendicular  erected  at  the  middle  point  of 
the  line.]  (§42.) 

Therefore,  DG,  EH,  and  FK  meet  in  the  common 
point  0. 

137.  Cor.  The  poi7it  of  intersection  of  the  perpendicu- 
lars erected  at  the  middle  points  of  the  sides  of  a  triangle, 
is  equally  distant  from  the  vertices  of  the  triangle. 


64 


PLANE   GEOMETRY.  —  BOOK   I. 


Proposition  LII.     Theorem. 

138.    The  perpendiculars  from  the  vertices  of  a  triangle  to 
the  opposite  sides  meet  in  a  common  point. 


G 

Let  AD,  BE,  and  CF  be  the  perpendiculars  from  the  ver- 
tices of  the  triangle  ABC  to  the  opposite  sides. 

To  prove  that  AD,  BE,  and  CF  meet  in  a  common  point. 

Through  A,  B,  and  C,  draw  HK,  KG,  and  GH,  parallel, 
respectively,  to  BC,  CA,  and  AB. 

Then  since  AD  is  perpendicular  to  BC,  it  is  also  per- 
pendicular to  HK. 

[A  straight  line  perpendicular  to  one  of  two  parallels  is  perpen- 
dicular to  the  other.]  (§56.) 

Now  since  ABCH  and  ACBK  are  parallelograms, 

AH  =  BC,  and  AK  =  BC. 
[The  opposite  sides  of  a  parallelogram  are  equal.]  (§  104.) 

Whence,  AH  =  AK,  and  A  is  the  middle  point  of  HK. 
Then  AD  is  perpendicular  to  HK  at  its  middle  point. 
In  like  manner,  BE  and  CF  are  the  perpendiculars  erected 
at  the  middle  points  of  KG  and  GH,  respectively. 
Hence,  AD,  BE,  and  CF  meet  in  a  common  point  0. 

[The  perpendiculars  erected  at  the  middle  points  of  the  sides  of 
a  triangle  mee*  in  a  common  point.]  (§  136.) 

139.  Def.  a  median  of  a  triangle  is  a  line  drawn  from 
any  vertex  to  the  middle  point  of  the  opposite  side. 


RECTILINEAR  FIGURES.  65 

Proposition-  LIII.     Theorem. 
140.    The  medians  of  a  triangle  meet  in  a  common  point. 
C 


Let  JD,  BE,  and  CF  be  the  medians  of  the  triangle  ABC. 
To  prove  that  AD,  BE,  and  CF  meet  in  a  common  point. 

Let  AD  and  BE  intersect  at  0. 

Let  G  and  H  be  the  middle  points  of  OA  and  OB, 
respectively,  and  draw  ED,  GH,  EG,  and  DH. 

Then  since  ED  bisects  AC  and  BC,  it  is  parallel  to  AB, 
and  equal  to  ^  AB. 

[The  line  joining  the  middle  points  of  two  sides  of  a  triangle  is 
parallel  to  the  third  side,  and  equal  to  one-half  of  it.]  (§  130.) 

In  like  manner,  since  GH  bisects  OA  and  OB,  it  is  par- 
allel to  AB,  and  equal  to  ^  AB. 

Therefore,  ED  and  GH  are  equal  and  parallel,  and 
EDHG  is  a  parallelogram. 

[If  two  sides  of  a  quadrilateral  are  equal  and  parallel,  the  figure 
is  a  parallelogram.]  (§  109.) 

Then  GD  and  EH  bisect  each  other  at  0. 

[The  diagonals  of  a  parallelogram  bisect  each  other,]  (§  110.) 

But  by  hypothesis,  G  is  the  middle  point  of  OA. 

Hence,  AG  =r  OG  =  OD,  and  OA  =  §  AD. 

That  is,  BE  intersects  AD  two-thirds  the  way  from  A 
to  BG. 

In  like  manner,  it  may  be  proved  that  CF  intersects  AD 
two-thirds  the  way  from  A  to  BC. 

Hence,  AD,  BE,  and  CF  meet  in  the  common  point  0. 


66  PLANE  GEOMETRY.  —  BOOK  I. 


LOCI. 

141.  Def.  If  a  series  of  points,  all  of  which  satisfy 
a  given  condition,  lie  in  a  certain  line,  that  line  is  called 
the  locus  of  the  points. 

For  example,  every  point  which  satisfies  the  condition  of 
being  equally  distant  from  the  extremities  of  a  straight 
line,  lies  in  the  perpendicular  erected  at  the  middle  point 
of  the  line  (§  42). 

Hence,  the  perpendicular  erected  at  the  middle  point  of  a 
straight  line  is  the  locus  of  points  which  are  equally  dis- 
tant from  the  extremities  of  the  line. 

Again,  every  point  which  satisfies  the  condition  of  being 
within  an  angle,  and  equally  distant  from  its  sides,  lies  in 
the  bisegtor  of  the  angle  (§  100). 

Hence,  the  bisector  of  an  angle  is  the  locus  of  points  which 
are  ivithin  the  angle,  and  equally  distant  from  its  sides. 

EXERCISES. 

60.  What  is  the  locus  of  points  at  a  given  distance  from  a  given 
straight  line  ? 

61.  What  is  the  locus  of  points  equally  distant  from  a  pair  of 
intersecting  straight  lines  ? 

62.  What  is  the  locus  of  points  equally  distant  from  a  pair  of 
parallel  straight  lines  ? 

63.  Two  isosceles  triangles  are  equal  when  the  base  and  vertical 
angle  of  one  are  equal  respectively  to  the  base  and  vertical  angle  of 
the  other. 

64.  If  from  any  point  in  the  base  of  an  isosceles  triangle  parallels 
to  the  equal  sides  be  drawn,  the  perimeter  of  the  parallelogram 
formed  is  equal  to  the  sum  of  the  equal  sides  of  the  triangle. 

65.  If  an  exterior  angle  be  formed  at  the  vertex  of  an  isosceles 
triangle,  its  bisector  is  parallel  to  the  base. 

66.  The  medians  drawn  from  the  extremities  of  the  base  of  an 
isosceles  triangle  are  equal. 

67.  State  and  prove  the  converse  of  Ex.  54,  p.  60. 

68.  The  diagonals  of  a  rhombus  bisect  its  angles. 


KECTILINEAR  FIGURES.  67 

69.  If  from  the  vertex  of  one  of  the  equal  angles  of  an  isosceles 
triangle  a  perpendicular  be  drawn  to  the  opposite  side,  it  makes  with 
the  base  an  angle  which  is  equal  to  one-half  the  vertical  angle  of  the 
triangle. 

70.  Every  point  within  an  angle,  and  not  in  the  bisector,  is  un- 
equally distant  from  the  sides  of  the  angle. 

71.  If  from  any  point  in  the  bisector  of  an  angle  a  parallel  to  one 
of  the  sides  be  drawn,  the  bisector,  the  parallel,  and  the  remaining 
side  form  an  isosceles  triangle. 

72.  If  the  bisectors  of  the  equal  angles  of  an  isosceles  triangle 
meet  the  equal  sides  at  D  and  E,  prove  that  BE  is  parallel  to  the 
base  of  the  triangle. 

73.  If  the  exterior  angles  at  the  vertices  A  and  ^  of  a  triangle 
ABC  are  bisected  by  lines  which  meet  at  D,  prove  that 

Z  ADB  =  90°  -  ^  C. 

74.  If  at  any  point  D  in  one  of  the  equal  sides  AB  of  the  isosceles 
triangle  ABC^  BE  is  drawn  perpendicular  to  the  base  BC  meeting 
CA  produced  at  E,  prove  that  the  triangle  ABE  is  isosceles. 

75.  From  C,  one  of  the  extremities  of  the  base  BC  of  an  isosceles 
triangle  ABC^  a  line  is  drawn  meeting  BA  produced  at  D,  making 
AB  =  AB.     Prove  that  CB  is  perpendicular  to  BC.     (§  91.) 

76.  If  the  non-parallel  sides  of  a  trapezoid  are  equal,  its  diagonals 
are  also  equal. 

77.  If  ABC  is  a  re-entrant  angle  of  the  quadrilateral  ABCB, 
prove  that  the  angle  ABC,  exterior  to  the  figure,  is  equal  to  the  sum 
of  the  interior  angles  A,  B,  and  C 

78.  If  a  diagonal  of  a  quadrilateral  bisects  two  of  its  angles,  it  is 
perpendicular  to  the  other  diagonal. 

79.  If  two  lines  are  cut  by  a  third  so  as  to  make  the  sum  of  the 
interior  angles  on  the  same  side  of  the  secant.line  less  than  two  right 
angles,  the  lines  will  meet  if  sufficiently  produced. 

80.  If  exterior  angles  be  formed  at  two  vertices  of  a  triangle, 
their  bisectors  will  intersect  on  the  bisector  of  the  interior  angle  at 
the  third  vertex.     (§  134.) 

81.  In  a  quadrilateral  ABCB,  the  angles  ABB  and  CAB  are 
equal  to  ACB  and  BBA  respectively;  prove  that  BC  is  parallel 
to  AB. 

82.  ABCB  is  a  trapezoid  whose  parallel  sides  AB  and  BC  are 
perpendicular  to  CB.  If  E  is  the  middle  point  of  AB,  prove  that 
EC  =  EB. 


68  PLANE   GEOMETRY.  —  BOOK  I. 

83.  State  and  prove  the  converse  of  Prop.  XLV. 

84.  State  and  prove  the  converse  of  Ex.  65,  p.  66. 

85.  Prove  Prop.  XXVI.  by  drawing  through  B  a  parallel  to  AC. 

86.  Prove  Prop.  XXX.  by  drawing  CD  to  the  middle  point  of  AB. 

87.  Prove  Prop.  XXXI.  by  drawing  CD  so  as  to  bisect  Z  ACB. 

88.  The  middle  point  of  the  hypotenuse  of  a  right  triangle  is 
equally  distant  from  the  vertices  of  the  triangle. 

89.  The  bisectors  of  the  angles  .of  a  rectangle  form  a  square. 

90.  If  D  is  the  middle  point  of  the  side  BC  of  the  triangle  ABC, 
and  BE  and  CF  are  the  perpendiculars  from  B  and  C  to  AD,  pro- 
duced if  necessary,  prove  that  BE  =  CF.     • 

91.  The  angle  at  the  vertex  of  an  isosceles  triangle  ABC  is  equal 
to  twice  the  sum  of  the  equal  angles  B  and  C.  If  CD  be  drawn  per- 
pendicular to  BC,  meeting  BA  produced  at  D,  prove  that  the  triangle 
ACD  is  equilateral. 

92.  The  bisector  of  the  vertical  angle  A  of  an  equilateral  triangle 
ABC  is  produced  to  D,  so  that  AD  =-  AB.  If  BD  and  CD  be 
drawn,  prove  that  Z  BDC  is  30°  or  150°,  according  as  D  lies  above 
or  below  the  base. 

93.  If  the  angle  B  of  the  triangle  ABC  is  greater  than  the  angle 
C,  and  BD  be  drawn  to  AC  making  AD  =  AB,  prove  that 

Z  ADD  =  i(B+C),  and  Z  CBD  ^  i  (B  -  C). 

94.  The  sum  of  the  lines  drawn  from  any  point  within  a  triangle 
to  the  vertices  is  greater  than  the  half-sum  of  the  three  sides.    (§  61.) 

95.  The  sum  of  the  lines  drawn  from  any  point  within  a  triangle 
to  the  vertices  is  less  than  the  sum  of  the  three  sides. 

96.  How  many  sides  are  there  in  the  polygon  the  sum  of  whose 
interior  angles  exceeds  the  sum  of  its  exterior  angles  by  540°  ? 

97.  If  D,  E,  and  jP  are  points  on  the  sides  AB,  BC,  and  CA  of  an 
equilateral  triangle  ABC,  such  that  AD  ^  BE  =  CF,  prove  that  the 
figure  DEF  is  an  equilateral  triangle. 

98.  If  E,  F,  G,  and  H  are  points  on  the  sides  AB,  BC,  CD,  and 
DA  of  a  parallelogram  ABCD,  such  that  AE  =  CG  and  BF=  DH, 
prove  that  the  figure  EFGH  is  a  parallelogram. 

99.  If  E,  F,  G,  and  H  are  points  on  the  sides  AB,  BC,  CD,  and 
DA  of  a  square  ABCD,  such  that  AE  =  BF  =CG  =  DH,  prove  that 
the  figure  EFGH  is  a  square. 


KECTILINEAR  FIGURES.  69 

100.  If  on  the  diagonal  BD  of  a  square  ABCD  a  distance  BE  is 
taken  equal  to  AB,  and  EF  is  drawn  perpendicular  to  BD,  meeting 
AD  at  F,  prove  that  AF  =  EF  =  ED. 

101.  Prove  the  theorem  of  §  127  by  drawing  lines  from  any  point 
within  the  polygon  to  the  vertices. 

102.  State  and  prove  the  converse  of  Ex.  68,  p.  66. 

103.  State  and  prove  the  converse  of  Prop.  XXXVIII. 

104.  State  and  prove  the  converse  of  Ex.  76,  p.  67. 

105.  If  AD  is  the  perpendicular  from  the  vertex  of  the  right 
angle  to  the  hypotenuse  of  the  right  triangle  ABC,  and  AE  is  the 
bisector  of  the  angle  A,  prove  that  /.DAE  is  equal  to  one-half  the 
difference  of  the  angles  B  and  C. 

106.  D  is  any  point  in  the  base  BC  of  an  isosceles  triangle  ABC. 
The  side  J.C  is  produced  below  C  to  E,  so  that  CE  =  CD,  and  DE  is 
drawn  meeting  AB  at  F.     Prove  that  Z.  AFE  =  3  Z  AEF. 

107.  If  ABC  and  ABD  are  two  triangles  on  the  same  base  and  on 
the  same  side  of  it,  such  that  AC  =BD  and  AD  =  BC,  and  AD  and 
BC  intersect  at  O,  prove  that  the  triangle  OAB  is  isosceles. 

108.  If  D-is  the  middle  point  of  the  side  AC  of  the  equilateral 
triangle  ABC,  and  DE  be  drawn  perpendicular  to  BC,  prove  that 
EC^i  BC. 

109.  If  in  the  parallelogram  ABCD,  E  and  F  are  the  middle 
points  of  the  sides  BC  and  AD,  prove  that  the  lines  AE  and  CF 
trisect  the  the  diagonal  BD. 

110.  If  ^D  is  the  perpendicular  from  the  vertex  of  the  right  angle 
to  the  hypotenuse  of  the  right  triangle  ABC,  and  E  is  the  middle 
point  of  BC,  prove  that  Z  DAE  is  equal  to  the  difference  of  the 
angles  B  and  C. 

111.  If  one  acute  angle  of  a  right  triangle  is  double  the  other,  the 
hypotenuse  is  double  the  shortest  leg. 

112.  If  AD  be  drawn  from  the  vertex  of  the  right  angle  to  the 
hypotenuse  of  the  right  triangle  ^jBO  so  as  to  make  Z  DAC  =  Z  C, 
it  bisects  the  hypotenuse. 

113.  If  D  is  the  middle  point  of  the  side  SC  of  the  triangle 
ABC,^roYethsitAD>i(iAB  + AC-BC).     (§62.) 


Note.     For  additional  exercises  on  Book  I.,  see  p.  221. 


BOOK  II. 

THE    CIRCLE. 


•DEFINITIONS. 

142.  A  circle  is  a  portion  of  a  plane  bounded  by  a  curve 
called   a  circuinference,  all   points   of 

which  are  equally  distant  from  a  point 
within,  called  the  centre ;  as  ABCD. 

Any  portion  of  the  circumference,  as 
AB,  is  called  an  arc. 

A  radius  is  a  straight  line  drawn 
from  the  centre  to  the  circumference ; 
as  OA. 

A  diameter  is.  a  straight  line  drawn  through  the  centre, 
having  its  extremities  in  the  circumference ;  as  AC. 

143.  It  follows  from  the  definition  of  §  142  that 
All  radii  of  a  circle  are  equal. 

Also,  all  its  diameters  are  equal,  since  each  is  the  sum  of 
two  radii. 

144.  Two  circles  are  equal  when  their  radii  are  equal. 
For  they  can  evidently  be  applied  one   to  the  other  so 

that  their  circumferences  shall  coincide  throughout. 

145.  Conversely,  the  radii  of  equal  circles  are  equal. 

146.  A  semi-circumference  is  an  arc  equal  to  one-half  the 
circumference ;  and  a  quadrant  is  an  arc  equal  to  one-fourth 
the  circumference. 

Concentric  circles  are  circles  having  the  same  centre. 

70 


THE   CIRCLE. 


n 


147.  A  chord  is  a  straight  line  joining  the  extremities  of 
an  arc ;  as  AB. 

The  arc  is  said  to  be  subtended  by  its 
chord. 

Every  chord  subtends  two  arcs  ;  thus 
the  chord  AB  subtends  the  arcs  AMB  and 
and  ACDB. 

When  the  arc  subtended  by  a  chord  is 
spoken  of,  that  arc  which  is  less  than  a 
semi-circumference  is  understood,  unless   the   contrary  is 
specified. 

A  segment  of  a  circle  is  the  portion  included  between  an 
arc  and  its  chord ;  as  AMBN. 

A  semicircle  is  a  segment  equal  to  one-half  the  circle. 

A  sector  of  a  circle  is  the  portion  included  between  an 
arc  and  the  radii  drawn  to  its  extremities;  as  OCD. 

148.  A  central  angle  is  an  angle  whose  vertex  is  at  the 
centre,  and  whose  sides  are  radii ;  as  ^OC. 

An  inscribed  angle  is  an  angle  whose  ver- 
tex is  on  the  circumference,  and  whose 
sides  are  chords ;  as  ABC. 

An  angle  is  said  to  be  inscribed  in  a 
segment  when  its  vertex  is  on  the  arc  of 
the  segment,  and  its  sides  pass  through  the 
extremities  of  the  subtending  chord. 

Thus,  the  angle  B  is  inscribed  in  the  segment  ABC. 

149.  A^straight  line  is  said  to  touch,  or  be  tangent  to,  a 
circle  when  it  has  but  one  point  in  com- 
mon with  the  circumference ;  as  AB. 

In  such  a  case,  the  circle  is  said  to 
be  tangent  to  the  straight  line. 

The  common  point  is  called  the 
point  of  contact,  or  point  of  tangency. 

A  secant  is  a  straight  line  which 
intersects  the  circumference   in  two  points ;    as   CD. 


72 


PLANE   GEOMETRY.  —  BOOK  11. 


150.  Two  circles  are  said  to  be  tangent  to  each  other  when 
they  are  both  tangent  to  the  same  straight  line  at  the  same 
point. 

They  are  said  to  be  tangent  internally  or  externally 
according  as  one  circle  lies  entirely  within  or  entirely  with- 
out the  other. 

A  common  tangent  to  two  circles  is  a  straight  line  which 
is  tangent  to  both  of  them. 


151.  A  polygon  is  said  to  be  inscribed 
in  a  circle  when  its  vertices  lie  on  the 
circumference  ;  as  ABCD. 

In  such  a  case,  the  circle  is  said  to  be 
circumscribed  about  the  polygon. 

A  polygon  is  said  to  be  inscrlptible 
when  it  can  be  inscribed  in  a  circle. 

A  polygon  is  said  to  be  circu7nscribed 
about  a  circle  when  its  sides  are  tangent 
to  the  circle ;  as  EFGH. 

In  such  a  case,  the  circle  is  said  to  be 
inscribed  in  the  polygon. 


Proposition   I.     Theorem. 
152.   Every  diameter  bisects  the  circle  and  its  circumference. 

B 


Let  ^C  be  a  diameter  of  the  circle  ABCD. 

To  prove  that  AC  bisects  the  circle  and  its  circumference. 


THE   CIRCLE.  73 

Superpose  the  segment  ABC  upon  ADC,  by  folding  it 
over  about  AC  as  an  axis. 

Then,  the  arc  ABC  will  coincide  with  the  arc  ADC;  for 
otherwise  there  would  be  points  of  the  circumference 
unequally  distant  from  the  centre. 

Hence,  the  segments  ^J5C'  Siiid  ADC  coincide  throughout, 
and  are  equal. 

Therefore,  AC  bisects  the  circle  and  its  circumference. 


Proposition   II.     Theorem. 

153.   A  straight  line  cannot  intersect  a  circumference  in 
more  than  two  points. 

0 


MA  B  C     N 


Let  0  be  the  centre  of  a  circle,  and  MN  any  straight  line. 
To  prove  that  MJ^  cannot  intersect  the  circumference  in 
more  than  two  points. 

If  possible,  let  MN  intersect  the  circumference  in  thi-ee 
points.  A,  B,  and  C;  and  draw  OA,  OB,  and  DC. 

Then,  OA=OB=OC.  (§  143.) 

We  should  then  have  three  equal  straight  lines  drawn 
from  a  point  to  a  straight  line,  which  is  impossible.     (§  49.) 

Therefore,  MN  cannot  intersect  the  circumference  in 
more  than  two  points. 

EXERCISES. 

1.  What  is  the  locus  of  points  at  a  given  distance  from  a  given  point? 

2.  If  two  circles  intersect  each  other,  the  distance  between  their 
centres  is  greater  than  the  difference  of  their  radii.     (§  62.) 


74 


PLANE   GEOMETRY.  —  BOOK  II. 


Proposition   III.     Theorem. 

154.    In  equal  circles,  or  in  the  same  circle,  equal  central 
angles  intercept  equal  arcs  on  the  circumference. 


M  M' 

Let  C  and  C  be  the  centres  of  the  equal  circles  AMB  and 
A'M'B' ;  and  let        ZACB=A  A' C'B'. 
To  prove  arc  AB  =  arc  A'B'. 

Superpose  the  sector  ABC  upon  AB'C  in  such  a  way 
that  Z  C  shall  coincide  with  its  equal  AC. 

Now,  AC  =  A'C,  and  BC  =  B'C.  (§  145.) 

Whence,  the  point  A  will  fall  at  A',  and  B  at  B'. 

Then,  the  arc  AB  will  coincide  with  the  arc  A'B'  -,  for 
all  points  in  each  are  equally  distant  from  the  centre. 

Therefore,  arc  AB  =  arc  A'B'. 


Proposition  IV.     Theorem. 

155.    (Converse  of  Prop.  III.)     In  equal  circles,  or  in  the 
same  circle,  equal  arcs  are  intercepted  by  equal  central  angles. 


Let  C  and  C  he  the  centres  of  the  equal  circles  AMB 
and  A'M'B' ;  and  let  arc  AB  =  arc  A'B'. 


THE   CIRCLE. 


75 


To  prove  ^ACB^^ZA'  C'B'. 

Since  the  circles  are  equal,  we  may  superpose  AMB  upon 
A'M'B'  in  such  a  way  that  the  point  A  shall  fall  at  A'  \ 
the  centre  C  falling  at  C. 

Then  since  arc  AB  =  arc  A'B\  the  point  B  will  fall  at  B'. 

Whence,  the  radii  AC  and  BC  will  coincide  with  A'C 
and  B'C,  respectively.  (Ax.  5.) 

Therefore,  Z.ACB  will  coincide  with  Z  A  C'B'. 

Whence,  Z  ACB  =  Z  ^'C'^'. 

156.  ScH.  In  equal  circles,  or  in  the  same  circle,  the 
greater  of  two  central  angles  intercepts  the  greater  arc  on 
the  circumference ;  a^nd  conversely.  • 


Proposition  V.     Theorem. 

157.    In  equal  circles,  or  in  the  same  circle,  equal  chords 
subtend  equal  arcs. 


In  the  equal  circles  AMB  and  A'M'B',  let 

chord  AB  =  chord  A'B". 
To  prove  arc  AB  =  arc  A'B'. 

Let  C  and   C  be  the  centres  of  the  circles;  and  draw 
AC,  BC,  A'C,  and  B'C. 

Then  in  the  triangles  ABC  and  A'B'C,  by  hypothesis, 

AB  =  A'B'. 
Also,  AC  =  A'C,  and  BC  =  B'C  (§  145.) 

Therefore,  A  ABC  =  A  A'B'C.  (§  69.> 

Whence,  ZC  =  ZC  (§  66.) 

Therefore,  arc  AB  =  arc  A'B'.  '    (§  154.) 


76 


PLANE   GEOMETRY.  —  BOOK  II. 


Proposition  VI.     Theorem. 

158.    (Converse  of  Prop.  V.)     In  equal  circles,  or  in  the 
same  circle,  equal  arcs  are  subtended  by  equal  chords. 


In  the  equal  circles  AMB  and  A'M'B',  let 

arc  AB  =  arc  A'B'. 
To  prove  chord  AB  =  chord  A'B\ 

Since  the  circles  are  equal,  we  may  superpose  AMB  upon 
AM'B'  in  such  a  way  that  the  point  A  shall  fall  at  A'. 
Then  since  arc  AB  =  arc  A'B',  the  point  B  will  fall  at  B\ 
Therefore,  chord  AB  will  coincide  with  chord  A'B\ 

(Ax.  5.) 
Whence,  chord  AB  =  chord  A'B\ 


Proposition  VIL     Theorem. 

159.  In  equal  circles,  or  in  the  same  circle,  the  greater  of 
two  arcs  is  sicbterided  by  the  greater  chord ;  each  arc  being 
less  than  a  semircircumference. 


M  M' 

Let  AMB  and  A'M'B'  be  equal  circles. 


THE   CIRCLE. 


77 


Let  arc  AB  be  greater  than  arc  A'B' ;  each  arc  being  less 
than  a  semi-circumference. 

To  prove  chord  AB  >  chord  A'B'. 

Let  C  and  C"  be  the  centres  of  the  circles ;  and  draw  AC, 
BC,  AC,  and  B'C 

Then  in  the  triangles  ABC  and  A'B'C, 

AC  =  AC,  and  BC  =  B'C.  (§  145.) 

And  since  arc  AB  >  arc  A'B', 

AC>AC  (§  156.) 

Therefore,  chord  AB  >  chord  AB'.  (§  89.) 

Proposition  VIII.     Theorem. 

160.  (Converse  of  Prop.  VII.)  In  equal  circles,  or  in  the 
same  circle,  the  greater  of  two  chords  subtends  the  greater 
arc  ;  each  arc  being  less  than  a  semircircumference. 


In  the  equal  circles  AMB  and  AM'B',  let  chord  AB  be 
greater  than  chord  AB'. 

To  prove  arc  AB  >  arc  A'B '; 

each  arc  being  less  than  a  semi-circumference. 

Let  C  and  C  be, the  centres  of  the  circles;  and  draw  AC, 
BC,  AC,  and  B'C. 

Then  in  the  triangles  ABC  and  AB'C, 

AC  =  AC,  and  BC  =  B'C.  (§  145.) 

And  by  hypothesis,      AB>A'B'. 

Therefore,  ZC>ZC.  (§  90.) 

Whence,  arc  AB  >  arc  A'B'.  (§  156.) 


78 


PLANE   GEOMETRY.  —  BOOK  II. 


161.  ScH.  If  each  arc  is  greater  than  a  semi-circum- 
ference, the  greater  arc  is  subtended  by  the  less  chord; 
and  conversely  the  greater  chord  subtends  the  less  arc. 

Pkoposition  IX.     Theorem. 

162.  The  diameter  perpendicular  to  a  chord  bisects  the 
chord  and  its  subtended  arcs. 


In  the  circle  ADB,  let  the  diameter  CD  be  perpendicular 
to  the  chord  AB. 

To  prove  that  CD  bisects  AB,  and  the  arcs  ACB  and 
ADB. 

Let  0  be  the  centre  of  the  circle,  and  draw  OA  and  OB. 

Then,  OA  =  OB.  (§  143.) 

Hence,  the  triangle  OAB  is  isosceles. 

Therefore,  CD  bisects  AB,  and  the  angle  AOB.       (§  92.) 

Whence,  Z  AOC  =  Z  BOC. 

Therefore,  arc  AC  =  arc  BC.  (§  154.) 

But,  arc  CAD  =  arc  CBD.  (§  152.) 

Whence, 

arc  CAD  —  arc  ^C  =  arc  CBD  -  arc  BC. 
That  is,  arc  AD  =  arc  BD. 

Hence,  CD  bisects  AB,  and  the  arcs  ACB  and  ADB. 

163.    CoR.    The  j^erpendicular  erected  at  the  middle  point 
of  a  chord  passes  through  the  centre  of  the  circle,  and  bisects 
.    the  arcs  subtended  by  the  chord. 


THE   CIRCLE.  79 


EXERCISES. 

3.  The  diameter  which  bisects  a  chord  is  perpendicular  to  it  and 
bisects  its  subtended  arcs. 

4.  The  diameter  which  bisects  an  arc  bisects  its  chord  at  right 
angles. 

5.  The  straight  line  which  bisects  the  arcs  subtended  by  a  chord 
bisects  the  chord  at  right  angles. 

6.  The  straight  line  which  bisects  a  chord  and  its  subtended  arc 
is  perpendicular  to  the  chord. 

7.  The  perpendiculars  to  the  sides  of  an  inscribed  quadrilateral  at 
their  middle  points  meet  in  a  common  point. 

Proposition  X.     Theorem. 

164.    In  the  same  circle,  or  in  equal  circles,  equal  chords 
are  equally  distant  from  the  centre. 


Let  AB  and  CD  be  equal  chords  of  the  circle  ABB. 

To  prove  AB  and  CD  equally  distant  from  the  centre  0. 

Draw  OE  and   OF  perpendicular  to   AB   and    CD,  re- 
spectively, and  draw  OA  and  DC. 

Then  E  is  the  middle  point  of  AB,  and  F  of  CD.   (§  162.) 
Now  in  the  right  triangles  OAE  and  OCF, 
AE  =  CF, 
being  halves  of  equal  chords. 

Also,         •  OA  =  OC.  (§  143.) 

Therefore,  A  OAE  =AOCF.  (§  88.) 

Whence,  OE  =  OF.  (§  66.) 

Then  AB  and  CD  are  equally  distant  from  0. 


80  PLANE   GEOMETRY.  —  BOOK  II. 

Proposition  XI.     Theorem.     < 

165.    (Converse  of  Prop.  X.)  In  the  same  circle,  or  in  equal 
circles,  chords  equally  distant  from  the  centre  are  equal. 

JB 


Let  0  be  the  centre  of  the  circle  ABD;  and  draw  OE 
and  OF  perpendicular  to  AB  and  CD,  respectively. 
To  prove  that  if  OU  =  OF,  chord  AB  =  chord  CD. 

Draw  OA  and  0(7;  then  in  the  right  triangles  OAF  and. 
OCF, 

OA  =  OC.  (§  143.) 

And  by  hypothesis,       OF  =  OF. 

Therefore,  A  OAF  =  AOCF.  (§  88.) 

Whence,  AF  =  CF  (§  66.) 

But  F  is  the  middle  point  of  AB,  and  F  of  CD.  (§  162.) 
Therefore,      2  AF  =  2  CF,  or  AB  =  CD. 

Proposition   XII.     Theorem. 

166.    In  the  same  circle,  or  in  equal  circles,  the  less  of  fiao 
chords  is  at  the  greater  distance  from  the  centre. 


B 


In  the  circle  ABD,  let  chord  AB  be  less  than  chord  CD. 


THE  CIRCLE.  81 

Draw  OF  and  OG  perpendicular  to  AB  and  CD,  respec- 
tively. 

To  prove  0F>  OG. 

Since  chord  AB  <  chord  CD,  arc  AB  <  arc  CD.    (§  160.) 

Lay  off  arc  CE  =  arc  AB,  and  draw  CE. 

Then,  chord  CE  =  chord  AB.  (§  158.) 

Draw  Olf  perpendicular  to  CE,  intersecting  CD  at  K. 

Then  OH  =  OF.  (§  164.) 

But,  OH  >  OiT. 

And,  OK  >  OG^.  (§  45.) 

Whence,  OH,  or  its  equal  OF,  is  greater  than  OG. 

Proposition   XIII.     Theorem. 

167.  (Converse  of  Prop.  XII.)  In  tlie  same  circle,  or  in 
equal  circles,  if  Uvo  chords  are  unequally  distant  from,  the 
centre,  the  inore  remote  is  the  less. 


In  the  circle  ABD,  let  chord  AB  be  more  remote  from 
the  centre  0  than  chord  CD. 

To  prove  chord  AB  <i  chord  CD. 

Draw  OG  and  OH  perpendicular  to  AB  and  CD,  respec- 
tively, and  on  OG  lay  off  OK  =  OH. 

Through  K  draw  the  chord  EF  perpendicular  to  OK. 
Then,  chord  EF  =  chord  CD.  .(§  165.) 

Now,  chord  AB  is  parallel  to  chord  EF.       0  (§  54.) 

Whence,  arc  ^^  <  arc  EF.  , 

Therefore,  chord  AB  <  chord  EF.  (§  159.) 

Whence,  chord  AB  <  chord  CD. 


82 


PLANE   GEOMETRY.  —  BOOK   II. 


168.  Cor.  A  diameter  of  a  circle  is  greater  than  any 
other  chord ;  for  a  chord  which  passes  through  the  centre 
is  greater  than  any  chord  which  does  not.  (§  167.) 


Proposition  XIV.     Theorem. 

169.    A  straight  line  i^erpendicular  to  a  radius  of  a  circle 
at  its  extremity  is  tangent  to  the  circle. 


ADC  B 

Let  the  line  AB  be  perpendicular  to  the  radius"  OC  of  the 
circle  ^C,  at  its  extremity  C. 

To  prove  AB  tangent  to  the  circle. 

Let  OD  be  an^i  other  straight  line  drawn  from  0  to  AB. 
Then,  0D>  OC.  (§45.) 

Therefore,  the  point  D  lies  without  the  circle. 
Then  every  point  of  AB  except  C  lies  without  the  circle, 
and  AB  is  tangent  to  the  circle.  (§  149.) 

Proposition  XV.     Theorem. 

170.    (Converse  of  Prop.  XIV.)    A  tangent  to  a  circle  is 
perpendicular  to  the  radius  drawn  to  the  point  of  contact. 


AC  B 

Let  the  line  AB  be  tangent  to  the  circle  EC. 


1 


THE   CIRCLE.  83 

To  prove  that  AB  is  perpendicular  to  the  radius  00 
drawn  to  the  point  of  Gontact. 

If  JB  is  tangent  to  the  circle  at  0,  every  point  of  AB 
except  0  lies  without  the  circle. 

Then  OC  is  the  shortest  line  that  can  be  drawn  from  0 
to  AB. 

Therefore,  OC  is  perpendicular  to  AB.  (§  45.) 

171.  CoK.  A  line  perpendicular  to  a  tangent  at  its  point 
of  contact  passes  through  the  centre  of  the  circle. 

EXERCISES. 

8.  The  tangents  to  a  circle  at  the  extremities  of  a  diameter  are 
parallel. 

9.  If  two  circles  are  concentric,  any  two  chords  of  the  greater 
which  are  tangent  to  the  less  are  equal.     (§  1G5.) 

Proposition  XVI.     Theorem. 

172.  Two  parallels  intercept  equal  arcs  on  a  circumference. 
Case  I.    When  one  line  is  a  tangent  and  the  other  a  secant, 

E 


Let  AB  be  tangent  to  the  circle  CED  at  U;  and  let  CD 
be  a  secant  parallel  to  AB. 

To  prove  arc  OJE  =  arc  DU. 

Draw  the  diameter  EF. 

Then  JEF  is  perpendicular  to  AB.  (§  170.) 

It  is  therefore  perpendicular  to  OD.  (§  56.) 

Whence,                   arc  OF  =  arc  DF.  (§  162.) 


84  PLANE  GEOMETRY.  —  BOOK  II. 

•    Case  II.    When  both  lines  are  secants. 


In  the  circle  ABD,  let  AB  and  CD  be  parallel  secants. 
To  prove  arc  AC  =  arc  BD. 

Draw  EF  parallel  to  AB,  tangent  to  the  circle  at  G. 
Then  EF  is  parallel  to  CD.  (§  ^b.) 

Now,  arc  AG  =  arc  BG, 

and  arc  CjG  =  arc  DG.         (§  172,  Case  I.) 

Subtracting,  we  have 

arc  AC  =  arc  BD. 

Case  III.    When  both  lines  are  tangents. 

E 


In  the  circle  EF,  let  the  lines  AB  and  CD  be  parallel, 
and  tangent  to  the  circle  at  E  and  F,  respectively. 
To  prove  arc  EGF  =  arc  EHF. 

Draw  the  secant  GH  parallel  to  AB. 
Then  GH  \^  parallel  to  CD.  (§  55.) 

Now,  arc  ^G^  =  arc  EH, 

and  arc  EG  =  arc  FH.         (§  172,  Case  I.) 

Adding,  we  have  arc  EGF  =  arc  EHF. 


THE   CIRCLE.  86 

173.    Coio.    The  strcdght  line  joining  the  ;Qoints  of  contact 
of  two  parallel  tangents  is  a  diameter.  t 


Proposition  XVII.     Theorem. 

174.    The  two  tangents  to  a  circle  from  an  outside  point  are 
equal. 


Let  AB  and  AC  he  tangent  at  the  points  B  and  C,  re- 
spectively, to  the  circle  whose  centre  is  0. 
To  prove  AB  =  AC. 

Draw  OA,  OB,  and  OC. 

Then   OB  and   OC  are  perpendicular  to  AB  and  AC, 

respectively.  (§  170.) 
Now  in  the  right  triangles  OAB  and  OAC,  OA  is  common. 

Also,                               OB  =  OC.  (§  143.) 

Therefore,               A  OAB  =AOAC.  (§  88.) 

Whence,                        AB  =  AC.  (§  66.) 

175.  Cor.  From  the  eqnal  triangles  OAB  and  OAC,  we 
have 

Z  OAB  =  Z  OAC,  and  Z  AOB  =  Z  AOC. 

That  is,  the  line  joining  the  centre  of  a  circle  to  the  point 
of  intersection  of  two  tangents,  bisects  the  angle  between  the 
tangents,  and  also  bisects  the  angle  between  the  radii  drawn 
to  the  points  of  contact. 

Ex.  10.  The  straight  line  drawn  from  the  centre  of  a  circle  to 
the  point  of  intersection  of  two  tangents  bisects  at  right  angles 
the  chord  joining  their  points  of  contact. 


86 


PLANE   GEOMETRY. —BOOK  11. 


Proposition  XVIII.     Theorem. 

176.    If  two  circumferences  intersect,  the  straight  line  join- 
ing their  centres  bisects  their  common  chord  at  right  angles. 


Let  0  and  0'  be  the  centres  of  two  circles  whose  circum- 
ferences intersect  at  A  and  B\  and  draw  00'  and  AB. 
To  prove  that  00'  bisects  AB  at  right  angles. 

The  point  0  is  equally  distant  from  A  and  B.        (§  143.) 
In  like  manner,  0'  is  equally  distant  from  A  and  B. 
Therefore,  00'  bisects  AB  at  right  angles.  (§  43.) 

Proposition  XIX.     Theorem. 

177.    If  two  circles  are  tangent  to  each  other,  the  straight 
line  joining  their  centres  passes  through  their  point  of  contact. 


Let  0  and  0'  be  the  centres  of  two  circles,  which  are 
tangent  to  the  straight  line  AB  at  A  (§  150). 

To  prove  that  the  straight  line  joining  0  and  0'  passes 
through  A. 

Draw  the  radii  OA  and  O'A. 


THE   CIRCLE.  87 

Then  OA  and  (7 A  are  perpendicular  to  AB.  (§  170.) 

Hence,  OA  and  O'A  lie  in  the  same  straight  line.    (§  38.) 
But  only  one  straight  line  can  be  drawn  between  0  and  (/. 

(Ax.  5.) 
Therefore,  the  straight  line,  joining    0  and    0'  passes 
through  A. 

EXERCISES. 

11.  If  the  inscribed  and  circumscribed  circles  of  a  triangle  are 
concentric,  prove  that  the  triangle  is  equilateral. 

12.  Two  intersecting  chords  which  make  equal  angles  with  the 
diameter  passing  through  their  jioint  of  intersection  are  equal. 
(§  165.) 

,13.  In  an  inscribed  trapezoid,  the  non-parallel  sides  are  equal, 
and  also  the  diagonals.     (§158.) 

14.  If  AB  and  AC  are  the  tangents  from  the  point  A  to  the  circle 
whose  centre  is  O,  prove  that  Z  BAG  =  2  Z  OBC. 

ON   MEASUREMENT. 

178.  JRatio  is  the  relation  with  respect  to  magnitude 
which  one  quantity  bears  to  another  of  the  same  kind; 
and  is  expressed  by  writing  the  first  quantity  as  the  nume- 
rator, and  the  second  as  the  denominator,  of  a  fraction. 

Thus,  if  a  and  b  are  quantities  of  the  same  kind,  the 
ratio  of  a  to  b  is  expressed  y ;  it  may  also  be  expressed  a  :  b. 

179.  A  geometrical  magnitude  is  measured  by  finding  its 
ratio  to  another  magnitude  of  the  same  kind,  called  the 
unit  of  measure. 

Thus,  the  measure  of  the  line  AB,  A^ ' ' ' ^B 

referred  to  the  line  CD  as  the  unit, 

.    AB  C' 'D 

If  the  quotient  can  be  obtained  exactly  as  an  integer  or 
fraction,  the  result  is  called  the  numerical  measure  of  the 
given  magnitude. 

Thus,  if  CD  is  contained  4  times  in  AB,  the  numerical 
measure  of  AB,  with  respect  to  CD  as  the  unit,  is  4. 


88  PLANE   GEOMETRY.  —  BOOK   II. 

180.    Two  geometrical  magnitudes  of  the  same  kind  are 
said  to  be  commensurable  when  there 

is  another  magnitude  of  the  same  A^ ' ' ' ^B 

kind  which  is  contained  an  integral 

number  of  times  in  each,  C' ' ' '  D 

Thus,  if  the  line  EF  is  contained 


4  times  in  the  line  AB,  and  3  times  E^ ^F 

in  the   line  CD,  AB   and  CD 'are 
commensurable. 

Again,  two  lines  whose  lengths  are  2|  inches  and  3f  inches 
are  commensurable ;  for  the  unit  of  measure  ^^  inch  is  con- 
tained an  integral  number  of  times  in  each ;  i.e.,  55  times 
in  the  first  line,  and  76  times  in  the  second. 

181.  Two  geometrical  magnitudes  of  the  same  kind  are 
said  to  be  incommensurable  when  no  magnitude  of  the  same 
kind  can  be  found  which  is  contained  an  integral  number 
of  times  in  each. 

For  example^  let  AB  and  CD  be  two  lines  such  that 

CD 

Since  the  value  of  V^  can  only  be  obtained  approxi- 
mately as  a  decimal,  it  is  evident  that  no  line,  however 
small,  can  be  found  which  is  contained  an  integral  number 
of  times  in  each  line. 

Then  AB  and  CD  are  incommensurable. 

182.  Although  a  magnitude  which  is  incommensurable 
with  respect  to  the  unit  has,  strictly  speaking,  no  numerical 
measure  (§  179),  still  if  CD  is  the  unit  of  measure,  and 

AB         /-  - 

=  V2,  we  shall  speak  of  V2  as  the  numerical  meas- 

L/D 

ure  of  AB. 

183.  It  is  evident  from  the  above  that  the  ratio  of  two 
magnitudes  of  the  same  kind,  whether  commensurable  or 
incommensurable,  is  equal  to  the  ratio  of  their  numerical 
measures  when  referred  to  a  common  unit. 


3 


THE  CIRCLE.  89 

THE  METHOD   OF  LIMITS. 

184.  A  variable  quantity,  or  simply  a  variable,  is  a  quan- 
tity which  may  assume,  under  the  conditions  imposed  upon 
it,  an  indefinitely  great  number  of  different  values. 

185.  A  constant  is  a  quantity  which  remains  unchanged 
throughout  the  same  discussion. 

186.  A  limit  of  a  variable  is  a  constant  quantity,  the 
difference  between  which  and  the  variable  may  be  made 
less  than  any  assigned  quantity,  however  small,  without 
ever  becoming  zero. 

In  other  words,  a  limit  of  a  variable  is  a  fixed  quantity 
which  the  variable  may  approach  as  near  as  we  please,  but 
can  never  actually  reach. 

The  variable  is  said  to  approach  its  limit. 

187.  Suppose,  for  example,  that  a  point  moves  from  A 
towards  B  under  the  condition  that    j^  C     D  E  B 

it  shall  move,  during  successive  equal     ' « ' — i— » 

intervals  of  time,  first  from  A  to  C,  half-way  from  ^  to  ^ ; 
then  to  D,  half-way  from  C  to  ^  ;  then  to  ^,  half-way  from 
i>  to  ^ ;  and  so  on  indefinitely. 

In  this  case,  the  distance  from  A  to  the  moving  point  is  a 
variable  which  approaches  the  constant  value  AB  as  a  limit; 
for  the  distance  between  the  moving  point  and  B  can  be 
made  less  than  any  assigned  quantity,  however  small,  but 
cannot  be  made  actually  equal  to  zero. 

Again,  the  distance  from  B  to  the  moving  point  is  a  vari- 
able which  approaches  the  limit  0. 

As  another  illustration,  consider  the  series 

where  each  term  after  the  first  is  one-half  the  preceding. 

In  this  case,  by  taking  terms  enough,  the  last  term  may 
be  made  less  than  any  assigned  number,  however  small,  but 
cannot  be  made  actually  equal  to  0. 


90  PLANE   GEOMETRY.  —  BOOK  II. 


Then,  the  last  term  of  the  series  is  a  variable  which 
approaches  the  limit  0  when  the  number  of  terms  is  in- 
definitely increased. 

Again,  the  sum  of  the  first  two  terms  is  1^ ; 
the  sum  of  the  first  three  terms  is  1| ; 
the  sum  of  the  first  four  terms  is  1| ;  etc. 

In  this  case,  by  taking  terms  enough,  the  sum  of  the 
terms  may  be  made  to  differ  from  2  by  less  than  any  as- 
signed number,  however  small,  but  cannot  be  made  actually 
equal  to  2. 

Then,  the  sum  of  the  terms  of  the  series  is  a  variable 
which  approaches  the  limit  2  when  the  number  of  terms 
is  indefinitely  increased. 

188.  The  Theorem  of  Limits.  Jf  two  variables  are 
always  equal,  and  each  aj^proaches  a  limit,  the  limits  are 
equal.  ^ 

AM  G      B 

i 1 : I  ,  I 


A'       M'  B' 

I \ I 


Let  AM  and  A'M'  be  two  variables,  which  are  always 
equal,  and  approach  the  limits  AB  and  A'B',  respectively. 
To  prove  AB  =  A'B'. 

If  possible,  let  AB  be  greater  than  A'B',  and  lay  off 
AC  =  A'B'. 

Then  the  variable  AM  may  assume  values  between  AC 
and  AB,  while  the  variable  A'M'  is  restricted  to  values  less 
than  AC',  which  is  contrary  to  the  hypothesis  that  the 
variables  are  always  equal. 

Hence,  AB  cannot  be  greater  than  A'B'. 

In  like  manner,  it  may  be  proved  that  AB  cannot  be  less 
than  A'B'. 

Therefore,  AB  =  A'B'. 


I 


THE   CIRCLE.  91 

MEASUREMENT  OF  ANGLES. 

Proposition  XX.     Theorem. 

189.    In  the  same  circle,  or  in  equal  circles,  two  central 
angles  are  in  the  same  ratio  as  their  intercepted  arcs. 

Case  I.    When  the  arcs  are  commensurable  (§  180). 

B 


In  the  circle  ABC,  let  AOB  and  BOC  he  central  angles, 
intercepting  the  commensurable  arcs  AB  and  BC,  respec- 
tively. 

,p  /.AOB       iitgAB 

^^P^""^  -ZBOC  =  ^i^^BC' 

Let  AD  be  a  common  measure  of  the  arcs  AB  and  BC; 
and  suppose  it  to  be  contained  4  times  in  AB,  and  3  times 
in  BC. 

Drawing  radii  to  the  several  points  of  division,  /.AOB 
will  be  divided  into  4  parts,  and  /BOC  into  3  parts,  all  of 
which  parts  will  be  equal.    •  (§  155.) 


From  (1)  and  (2),  we  have 

/AOB  _^ygAB 
/  BOC  ~  3ire  BC 


(Ax.l.) 


92  PLANE   GEOMETRY.  —  BOOK   II. 

Case  II.    When  the  arcs  are  incommensurable  (§  181). 

B 


In  the  circle  ABC,  let  AOB  and  BOC  be  central  angles, 
intercepting  the  incommensurable  arcs  AB  and  BC,  respec- 
tively. 

rp  ZAOB      arcAB 

To  prove  = . 

^  ZBOC.     3iVG  BC 

Let  the  arc  AB  be  divided  into  any  number  of  equal 
parts,  and  let  one  of  these  parts  be  applied  to  the  arc  BC 
as  a  measure. 

Since  AB  and  BC  are  incommensurable,  a  certain  num- 
ber of  the  parts  will  extend  from  B  to  C,  leaving  a 
remainder  CC  less  than  one  of  the  parts. 

Draw  DC. 

Then  since  the  arcs  AB  and  BC^  are  commensurable,  we 
have 

ZAOB        arc  AB 


Z  BOC       arc  BC 


(§  189,  Case  I.) 


Now  let  the  number  of  subdivisions  of  the  arc  AB  be 
indefinitely  increased. 

Then  the  length  of  each  part  will  be  indefinitely  dimin- 
ished;  and  the  remainder  6'CV being  always  less  than  one 
of  the  parts,  will  approach  the  limit  0. 


Then,       ^^OB  ^^^^  approach  the  limit  ^A2^  , 
'       ZBOC  ^^  ZBOC' 

and,  will  approach  the  limit . 

arc^C"  .^^  slygBC 


1 


THE  CIRCLE.  93 

^^^'    /  i^i^ni  ^^^ 77777  ^^^  ^^^  variables  which  are 

ABOC  arc  BC 

always  equal,  and  approach  the  limits  and  ^^^    ^ 

respectively.  ^^^^  arc^C' 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 

Therefore,  ^^0^  ^  arc^^ 

Z.BOC       2iXcBC 

190.  ScH.  The  usual  unit  of  measure  for  arcs  is  the 
degree,  which  is  the  ninetieth  part  of  a  quadrant  (§  146). 

The  degree  of  arc  is  divided  into  sixty  equal  parts, 
called  minutes,  and  the  minute  into  sixty  equal  parts, 
called  seconds. 

If  the  sum  of  two  arcs  is  a  quadrant,  or  90°,  one  is  called 
the  complement  of  the  other ;  if  their  sum  is  a  semi-circum- 
ference, or  180°,  one  is  called  the  supplement  of  the  other. 

191.  CoR.  I.  By  §  154,  equal  central  angles,  in  the  same 
circle,  intercept  equal  arcs  on  the  circumference. 

Hence,  if  the  angular  magnitude  about  the  centre  of  a 
circle  be  divided  into  four  equal  parts,  each  part  will  inter- 
cept one-fourth  of  the  circumference ;  tliat  is, 

A  right  central  angle  intercepts  a  quadrant  on  the  circum- 
ference. 

192.  CoR.  II.  By  §  189,  a  central  angle  of  n  degrees 
bears  the  same  ratio  to  a  right  central  angle  as  its  inter- 
cepted arc  bears  to  a  quadrant. 

But  a  central  angle  of  n  degrees  is  ^^  of  a  right  central 
angle. 

Hence,  its  intercepted  arc  is  ^^  of  a  quadrant,  or  an  arc 
of  n  degrees. 

The  above  principle  is  usually  expressed  as  follows  : 

A  central  angle  is  measured  hy  its  intercepted  arc. 

This  means  simply  that  the  number  of  degrees  in  the 
angle  is  equal  to  the  number  of  degrees  in  its  intercepted 
arc. 


94 


PLANE   GEOMETRY.  —  BOOK  II. 


Proposition   XXI.     Theorem. 

193.    An    inscribed    angle    is    measured    hy   one-half   its 
intercepted  arc. 

Case  I.    When  one  side  of  the  angle  is  a  diameter. 


Let  ^C  be  a  diameter,  and  AB  a  chord,  of  the  circle  ABC. 
To  prove  that  ZBAC  is  measured  by  ^  arc  BC. 

Draw  the  radius  OB-,  then,  OA  =  OB.  (§  143.) 

AVhence,  ZB=ZA.  (§  91.) 

But,  ZBOC  =ZA-\-ZB.  (§83,1.) 

Therefore,  Z  BOC  =2  Z  A,  ot  ZA  =  ^Z  BOC. 
But,  ZBOC  is  measured  by  arc  BC.  (§  192.) 

Whence,  Z  A  is  measured  by  ^  arc  BC. 

Case  II.    When  the  centre  is  witliin  the  angle. 
A 


Let  AD  be  a  diameter  of  the  circle  ABC 
To  prove  that  the  inscribed  angle  BAC  is  measured  by 
I  arc  BC. 


THE   CIRCLE. 


95 


Now,  Z  BAD  is  measured  by  \  arc  BD, 

and  Z.  CAD  is  measured  by  ^  arc  CD. 

(§  193,  Case  I.) 

Therefore,  the  sum  of  the  angles  BAD  and  CAD  is  meas- 
ured by  one-half  the  «um  of  the  arcs  BD  and  CD, 

Hence,        Z  BAC  is  measured  by  ^  arc  BC. 

Case  III.    When  the  centre  is  without  the  angle. 


Let  AD  be  a  diameter  of  the  circle  ABC. 
To  prove  that  the  inscribed  angle  BAC  is  measured  by 
\  arc  BC. 

Now,  Z  5^D  is  measured  by  \  arc  BD, 

and  Z  CJD  is  measured  by  ^  arc  Ci>. 

(§  193,  Case  I.) 

Therefore,  the  difference  of  the  angles  BAD  and  CAD  is 
Measured  by  one-half  the  difference  of  the  arcs  BD  and  CD. 

Hence,        Z  BAC  is  measured  by  ^  arc  BC. 

194.  ScH.  As  explained  in  §  192,  this  proposition  means 
simply  that  the  number  of  degrees  in  an  inscribed  angle  is 
one-half  the  number  of  degrees  in  its  intercepted  arc. 

195.  Cor.  I.  All  angles  inscribed  in  the 
same  segment  are  equal. 

Thus,  if  the  angles  A,  B,  and  C  are 
inscribed  in  the  segment  ADE,  then  each 
angle  is  measured  by  ^  avc  DE.     (§  193.) 

Whence,  ZA=ZB=ZC. 


96 


PLANE   GEOMETRY.  —  BOOK  11. 


196.  Cor.  II.    An  angle  inscribed  in  a   semicircle   is   a 
right  angle. 

Let  ^C  be  a  diameter  of  the  circle 
ABD,  and  BAG  an  angle  inscribed  in 
the  semicircle  ABC, 

To  prove  BAC  a  right  angle. 

ABAC  is  measured  by  \  arc  BDC, 

(§  193.) 

But,  \  arc  BDC  is  a  quadrant. 
Whence,  Z  BAC  is  a  right  angle. 

Pkoposition  XXII.     Theorem. 

197.  The  angle  between  a  tangent  and  a  chord  is  meas- 
ured by  one-half  its  intercepted  arc. 


Let  AE  be  tangent  to  the  circle  BCD  at  B,  and  let  BC 
be  a  chord.  • 

To  prove  that  Z  ABC  is  measured  by  ^  arc  BC. 
.  Draw  the  diameter  BD. 

Then  BD  is  perpendicular  to  AE.  (§  170.) 

Now  since  a  right  angle  is  measured  by  one-half  a  semi- 
circumference, 

Z  ^j5i)  is  measured  by  |  arc  J5Ci>. 

Also,       Z  (7J5i)  is  measured  by  I  arc  Ci).  (§  193.) 

Hence, 
Z  ^^D  —  Z  CBD  is  measured  by  |  (arc  BCD  —  arc  Ci>). 

That  is,  Z  ^^(7  is  measured  by  ^  arc  ^C. 

In  like  manner,  Z  ^Z?C  is  measured  by  |  arc  J5Z>(7. 


THE  CIRCLE.  97 


Proposition  XXIII.     Theorem. 

198.  The  angle  between  two  chords,  intersecting  within 
the  circumference,  is  measured  by  one-half  the  sum  of  its 
intercepted  arc,  and  the  arc  intercepted  by  its  vertical  angle. 


In  the  circle  ABC,  let  the  chords  AB  and  CD  intersect 
atii'. 

To  prove  that 

Z  AEC  is  measured  by  ^  (arc  ^C  -|-  arc  BD). 

Draw  BC. 

Then,  Z  AEC  =  Z  B  +  Z  C.  (§  83,  1.) 

But,  Z  B  is  measured  by  ^  arc  AC, 

and  Z  C  is  measured  by  ^  arc  BD.  (§  193.) 

Whence, 

Z  AEC  is  measured  by  ^-  (arc  ^C  +  arc  BD). 

EXERCISES. 

15.  If,  in  the  figure  of  §  195,  the  arc  DE  is  -^j  of  the  circumfer- 
ence, how  many  degrees  are  there  in  the  angle  A  ? 

16.  If,  in  the  figure  of  §  197,  arc  BC  =  107°,  how  many  degrees 
are  there  in  the  angles  ABC  and  EBC  ? 

17.  If,  in  the  figure  of  §  197,  Z  ABC  ^  42°,  how  many  degrees 
are  there  in  the  arc  CD  ? 

18.  If,  in  the  above  figure,  arc  AD  =  94°,  and  Z  AEC  =  51°,  how 
many  degrees  are  there  in  the  arc  BC? 

19.  Prove  Prop.  XXII.  by  drawing  a  radius  perpendicular  to  BC. 
•  20.   Prove  Prop.  XXIII.  by  drawing  through  B  a  chord  parallel 

to  CD. 


98  PLANE   GEOMETRY.  —  BOOK   11. 

Proposition  XXIV.     Theorem. 

199.  The  angle  between  two  secants,  intersecting  without 
the  circumference,  is  measured  by  one-half  the  difference  of 
the  intercepted  arcs. 

f  I5> 


In  the  circle  ABC,  let  the  secants  AE  and  CE  inter- 
sect the  circumference  in  the  points  A  and  B,  and  C  and  D, 
respectively. 

To  prove  that  Z  ^  is  measured  by  ^  (arc  AC  —  arc  BB). 

Draw  BC. 

Then,  Z  ABC  =ZE-j-,Z  C.  (§  83,  1.) 

Whence,  Z  E  =:  Z  ABC  —  Z  C. 

But,       Z  ABC  is  measured  by  ^  arc  AC, 
and  Z  C  is  measured  by  ^  arc  BB.  (§  193.) 

Whence,     Z  ^  is  measured  by  ^  (arc  AC  —  arc  BB). 

EXERCISES. 

21.  If,  in  the  above  figure,  the  arcs  AC  and  BD  are  respectively 
two-ninths  and  one-sixteenth  of  tlie  circumference,  how  many  de- 
grees are  there  in  the  angle  E  ? 

22.  If,  in  the  above  figure,  arc  AC ^  117°,  and  ZC=  14°,  how 
many  degrees  are  there  in  the  angle  E  ? 

23.  If,  in  the  above  figure,  ^C  is  a  quadrant,  and  Z  E  =  39®,  how 
many  degrees  are  there  in  the  arc  BD  ? 

24.  Prove  Prop.  XXIV.  by  drawing  tlirough  B  a  chord  parallel 
to  CD. 


THE  CIRCLE.  99 

Proposition  XXV.     Theorem. 

200.    The  angle  between  a  secant  and  a  tangent  is  meas- 
ured by  one-half  the  difference  of  the  intercepted  arcs. 


Let  AE  be  tangent  to  the  circle  BDC  at  B ;  and  let  EO 
be  a  secant,  intersecting  the  circumference  in  the  points 
C  and  D. 

To  prove  that 

Z  E  is  measured  by  i  (arc  BFC  —  arc  BD). 

Draw  BC. 

Then,  Z  ABC  =  Z  E  +  Z  C.  (§  83,  1.) 

Whence,  ZE=  Z  ABC  —  Z  C. 

But,       ZABCis  measured  by  ^  arc  BFC.  (§  197.) 

And,  Z  C  is  measured  b}^  -J  arc  BD.  (§  193.) 

Whence,     Z  £^  is  measured  by  ^  (arc  BFC  —  arc  BD). 

EXERCISES. 

25.  If,  in  the  above  figure,  arc  BFC  =  197°,  and  arc  CD  =  16"^ 
how  many  degrees  are  there  in  the  angle  E  ? 

26.  If,  in  the  above  figure,  Z  ^=53°,  and  the  arc  BD  is  one-fifth 
of  the  circumference,  how  many  degrees  are  there  in  the  arc  BFC  ? 

27.  Prove  Prop.  XXV.  by  drawing  through  D  a  chord  parallel 
to  AE. 

28.  If  two  chords  intersect  at  right  angles  within  the  circumfer- 
ence of  a  circle,  the  sum  of  the  opposite  intercepted  arcs  is  equal  to 
a  semi-circumference. 


100  PLANE   GEOMETRY.  —  BOOK  II. 


Proposition   XXVI.     Theorem. 

201.    The  angle  between  two  tangents  is  measured  by  one- 
half  the  difference  of  the  intercepted  arcs. 


Let  AE  and  CE  be  tangent  to  the  circle  BDF  at  B  and 
D,  respectively. 
To  prove  that 

Z  ^  is  measured  by  ^  (arc  BFD  —  arc  BGD). 

Draw  BD. 

Then,  Z  ABD  =  ZE  +  Z  BDE.  (§  83,  1.) 

Whence,  ZE  =  Z  ABD  -  Z  BDE. 

But,  Z  ABD  is  measured  by  ^  arc  BFD, 

and  Z  BDE  is  measured  by  J  arc  ^G^D.        (§  197.) 

Whence, 

Z  j^  is  measured  by  ^  (arc  BFD  —  arc  BGD). 

202.   Cor.   We  have  from  the  above  figure, 
i  (arc  BFD  —  arc  BGD) 

=  I  (360°  -  arc  BGD  -  arc  ^6^Z>) 

=  1  (360°-2arc^(?i>) 

=  180°  —  arc  j^G^Z). 
Then,  Z.E  is,  measured  by  180°  —  arc  BGD. 
Hence,  the  angle  between  two  tangents  is  measured  by  the 
supplement  of  the  smaller  of  the  two  intercepted  arcs. 


THE  CIRCLE.  101 


EXERCISES. 


29.  If,  in  the  figure  of  §  201,  the  arc  BFD  is  thirteen-sixteenths 
of  the  circumference,  how  many  degrees  are  there  in  the  angle  E  ? 

30.  If  AB  and  AC  are  two  chords  of  a  circle  malcing  equal  angles 
with  the  tangent  at  A,  prove  that  AB  =  AC. 

31.  From  a  given  point  within  a  circle  and  not  coincident  with 
the  centre,  not  more  than  two  equal  straight  lines  can  be  drawn  to 
the  circumference.     (§  163.) 

32.  The  sum  of  two  opposite  sides  of  a  circumscribed  quadri- 
lateral is  equal  to  the  sum  of  the  other  two  sides.     (§  174.) 

33.  In  a  circumscribed  trapezoid,  the  straight  line  joining  the 
middle  points  of  the  non-parallel  sides  is  equal  to  one-fourth  the 
perimeter  of  the  trapezoid.     (§  132.) 

34.  If  the  opposite  sides  of  a  circumscribed  quadrilateral  are 
parallel,  the  figure  is  a  rhombus  or  a  square. 

35.  If  tangents  be  drawn  to  a  circle  at  the  extremities  of  any 
pair  of  diameters  which  are  not  perpendicular  to  each  other,  the 
figure  formed  is  a  rhombus. 

36.  If  the  angles  of  a  circumscribed  quadrilateral  are  right  angles, 
the  figure  is  a  square. 

37.  If  two  circles  are  tangent  to  each  other  at  the  point  A,  the 
tangents  to  them  from  any  point  in  the  common  tangent  which 
passes^  through  A  are  equal.     (§  174.) 

38.  If  two  circles  are  tangent  to  each  other  externally  at  the 
point  A,  the  common  tangent  wh^ch  passes  through  A  bisects  the 
other  two  common  tangents. 

39.  The  bisector  of  the  angle  between  two  tangents  to  a  circle 
passes  through  the  centre. 

40.  The  bisectors  of  the  angles  of  a  circumscribed  quadrilateral 
pa^s  through  a  common  point. 

41.  If  AB  is  one  of  the  non-parallel  sides  of  a  trapezoid  circum- 
scribed about  a  circle  whose  centre  is  C,  prove  that  ACS  is  a  right 
angle. 

42.  Three  consecutive  sides  of  an  inscribed  quadrilateral  subtend 
arcs  of  82°,  99°,  and  67°  respectively.  Find  each  angle  of  the  quad- 
rilateral in  degrees,  and  the  angle  between  its  diagonals. 

43.  An  angle  inscribed  in  a  segment  greater  than  a  semicircle  is 
acute  ;  and  an  angle  inscribed  in  a  segment  less  than  a  semicircle 
is  obtuse. 


102  PLANE   GEOMETRY.— BOOK  11. 

44.  The  opposite  angles  of  an  inscribed  quadrilateral  are  supple- 
mentary. 

45.  Prove  Prop.  YI.  by  drawing  radii  to  the  extremities  of  the 
chords. 

46.  Prove  the  theorem  of  §  168  by  drawing  radii  to  the  extremi- 
ties of  the  chord. 

47.  Prove  the  theorem  of  §  202  by  drawing  radii  to  the  points  of 
contact  of  the  tangents. 

48.  Prove  Prop.  XIII.  by  Beductio  ad  Absurdum. 

49.  If  any  number  of  angles  are  inscribed  in  the  same  segment, 
their  bisectors  pass  through  a  common  point.     (§  193.) 

50.  Two  chords  perpendicular  to  a  third  chord  at  its  extremities 
are  equal. 

51.  If  two  opposite  sides  of  an  inscribed  quadrilateral  are  equal 
and  parallel,  the  figure  is  a  rectangle. 

52.  If  the  diagonals  of  an  inscribed  quadrilateral  intersect  at  the 
centre  of  the  circle,  the  figure  is  a  rectangle. 

53.  The  circle  described  on  one  of  the  equal  sides  of  an  isosceles 
triangle  as  a  diameter,  bisects  the  base.     (§  196.) 

54.  If  a  tangent  be  drawn  to  a  circle  at  the  extremity  of  a  chord, 
the  middle  point  of  the  subtended  arc  is  equally  distant  from  the 
chord  and  from  the  tangent. 

55.  If  the  sides  AB,  BC,  and  CD  of  an  inscribed  quadrilateral 
subtend  arcs  of  99°,  106°,  and  78°  respectively,  and  the  sides  BA 
and  CD  produced  meet  at  E,  and  the  sides  AD  and  BC  at  F,  find 
the  niunber  of  degrees  in  the  angles  AED  and  AFB. 

56.  If  O  is  the  centre  of  the  circumscribed  circle  of  a  triangle 
ABC,  and  OD  is  drawn  perpendicular  to  BC,  prove  that 

Z  BOD  =  ZA. 

57.  If  D,  E,  and  F  are  the  points  of  contact  of  the  sides  AB, 
BC,  and  CA  of  a  triangle  circumscribed  about  a  circle,  prov«  that 

ZDEF=9(P-iA. 

58.  If  the  sides  AB  and  BC  of  an  inscribed  quadrilateral  ABCD 
subtend  arcs  of  69°  and  112°,  and  the  angle  AED  between  the 
diagonals  is  87°,  how  many  degrees  are  there  in  each  angle  of  the 
quadrilateral  ? 

59.  If  any  number  of  parallel  chords  be  drawn  in  a  circle,  their 
middle  points  lie  in  the  same  straight  line. 

,  60.   What  is  the  locus  of  the  middle  points  of  a  system  of  parallel 
chords  in  a  circle  ? 


THE   CIRCLE.  103 

61.  What  is  the  locus  of  the  middle  points  of  a  system  of  chords 
of  given  length  in  a  circle  ? 

62.  If  two  circles  are  tangent  to  each  other,  any  straight  line 
drawn  through  their  point  of  contact  subtends  arcs  of  the  same 
number  of  degrees  on  their  circumferences.     (§  197.) 

63.  If  a  straight  line  be  drawn  through  the  point  of  contact  of  two 
circles  which  are  tangent  to  each  other,  terminating  in  their  circum- 
ferences, the  radii  drawn  to  its  extremities  are  parallel. 

64.  If  a  straight  line  be  drawn  through  the  point  of  contact  of  two 
circles  which  are  tangent  to  each  other,  terminating  in  their  circum- 
ferences, the  tangents  at  its  extremities  are  parallel. 

65.  If  the  sides  AB  and  DC  of  an  inscribed  quadrilateral  be 
produced  to  meet  at  E,  prove  that  the  triangles  ACE  and  BDE, 
and  also  the  triangles  ADE  and  BCE,  are  mutually  equiangular. 

66.  The  sum  of  the  angles  subtended  at  the  centre  of  a  circle  by 
two  opposite  sides  of  a  circumscribed  quadrilateral  is  equal  to  two 
right  angles. 

67.  If  a  circle  be  described  on  the  radius  of  another  circle  as  a 
diameter,  any  chord  of  the  greater  passing  through  the  point  of 
contact  of  the  circles  is  bisected  by  the  circumference  of  the  smaller. 
(§  196.) 

68.  If  a  circle  is  inscribed  in  a  right  triangle,  the  sum  of  its 
diameter  and  the  hypotenuse  is  equal  to  the  sum  of  the  legs. 

69.  If  the  sides  AB  and  CD  of  a  quadrilateral  inscribed  in  a  circle 
make  equal  angles  with  the  diameter  passing  through  their  point  of 
intersection,  prove  that  AB  =  CD.     (§  165^ 

70.  If  the  angles  A,  B,  and  C  of  a  circumscribed  quadrilateral 
ABCD  are  128°,  67°,  and  112°,  and  the  sides  AB,  BC,  CD,  and  DA 
are  tangent  to  the  circle  at  the  points  E,  F,  G,  and  II,  find  the 
number  of  degrees  in  each  angle  of  the  quadrilateral  EFGH. 

TL.  The  least  chord  which  can  be  drawn  through  a  given  point 
within  a  circle  is  perpendicular  to  the  diameter  passing  through 
that  point.     (§165.) 

72.  If  D,  E,  and  F  are  the  middle  points  of  the  arcs  subtended 
by  the  sides  AB,  BC,  and  CA  of  an  inscribed  triangle,  prove  that 
the  sum  of  the  angles  ADB,  BEC,  and  CFA  is  equal  to  four  right 
angles. 

Note.     For  additional  exercises  on  Book  II.,  see  p.  222. 


104 


PLANE   GEOMETRY.  —  BOOK  II. 


PROBLEMS  IN  CONSTRUCTION. 

Proposition  XXVII.     Problem. 

203.    At  a  given  point  in  a  straight  line  to  erect  a  perpen- 
dicular to  that  line. 

First  Method. 


A    D 


E     B 


Let  C  be  the  given  point  in  the  straight  line  ^^. 
To  erect  a  perpendicular  to  AB  at  C. 

Lay  off  the  equal  distances  CD  and  CE. 

With  D  and  E  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  F,  and  draw  CF. 

Then  CF  is  perpendicular  to  AB  at  C. 

For  since  C  and  F  are  each  equally  distant  from  D  and 
E,  CF  is  perpendicular  to  DE  at  its  middle  point.     (§  43.) 

Second  Method. 


^ 


0..'' 


x>\. 


/C 


Let  C  be  the  given  point  in  the  straight  line  AB. 
To  erect  a  perpendicular  to  AB  at  C. 


T^E  CIRCLE.  105 

With  any  point  0,  without  the  line  AB,  as  a  centre,  and 
the  distance  OC  as  a  radius,  describe  a  circumference  inter- 
secting AB  at  C  and  D. 

Draw  the  diameter  DE ;  also,  draw  CE. 

Then  CE  is  perpendicular  to  AB  at  C. 

For  /.  DCEj  being  inscribed  in  a  semicircle,  is  a  right 
angle.  (§  196.) 

Whence,  CE  is  perpendicular  to  CD. 

Note.  The  second  method  of  construction  is  preferable  when  the 
given  point  is  near  the  end  of  the  line. 


Proposition  XXVIII.     Problem. 

204.    From  a  given  point  without  a  straight  line  to  draw 
a  perpendicular  to  that  line. 


C 


/T 


Let  C  be  the  given  point  without  the  straight  line  AB. 
To  draw  a  perpendicular  from  C  to  AB. 

With  C  as  a  centre,  and  with  any  convenient  radius, 
describe  an  arc  intersecting  AB  at  D  and  E. 

With  D  and  E  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  F,  and  draw  CF. 

Then  CF  is  perpendicular  to  AB. 

For  since  C  and  F  are  each  equally  distant  from  D  and 
E,  CF  is  perpendicular  to  DE  at  its  middle  point.     (§  43.) 


Ex.  73.   Given  the  base  and  altitude  of  an  isosceles  triangle,  to 
construct  the  triangle. 


106  PLANE   GEOMETRY.  —  BOOK  II. 

Proposition  XXIX.     Problem. 
205.    To  bisect  a  given  straight  line. 


1    * 

Let  AB  be  the  given  straight  line. 
To  bisect  AB, 

With  A  and  B  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  C  and  D. 

Draw  CD  intersecting  AB  at  E. 

Then  E  is  the  middle  point  of  AB. 

For  since  C  and  D  are  each  equally  distant  from  ^  and 
B,  CD  is  perpendicular  to  AB  at  its  middle  point.      (§  43.) 

Proposition  XXX.     Problem. 
206.    To  bisect  a  given  arc. 


Let  ABhe  the  given  arc. 


THE   CIRCLE.  107 

To  bisect  the  arc  AB. 

With  A  and  B  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  C  and  D. 

Draw  CD  intersecting  the  arc  AB  at  E. 

Then  E  is  the  middle  point  of  the  arc  AB. 

For,  draw  the  chord  AB. 

Then  CD  is  perpendicular  to  the  chord  AB  at  its  middle 
point.  (§  43.) 

Whence,  CD  bisects  the  arc  AB.  (§163.) 

Proposition  XXXI.     Problem. 
207.    To  bisect  a  given  angle. 


Let  AOB  be  the  given  angle. 
To  bisect  Z  ^0^5. 

With  0  as  a  centre,  and  with  any  convenient  radius, 
describe  an  arc  intersecting  OA  at  C,  and  OB  at  D. 

With  C  and  D  as  centres,  and  with  equal  radii,  describe 
arcs  intersecting  at  E  ;  and  draw  OE. 

Then  OE  bisects  ZAOB. 

For  since  0  and  E  are  each  equally  distant  from  C  and  D, 
OE  bisects  the  arc  CD  at  F.  (§  206.) 

Whence,  Z  COi^  =  Z  i)Oi<^.  (§  155.) 

That  is,  OE  bisects  Z  ^07?. 

208.  ScH.  By  repetitions  of  the  above  construction,  an 
angle  may  be  divided  into  4,  8,  16,  etc.,  equal  parts. 


108  PLANE   GEOMETRY.— BOOK   II. 


Proposition  XXXII.     Problem. 

209.    With  a  given  vertex  and  a  given  side,  to  construct  an 
angle  equal  to  a  given  angle. 


Let  0  be  the  given  vertex,  OA  the  given  side,  and  0'  the 
given  angle. 

To  construct,  with  0  as  a  vertex  and  OA  as  a  side,  an 
angle  equal  to  0\ 

With  (y  as  a  centre,  and  with  any  convenient  radius, 
describe  an  arc  intersecting  the  sides  of  the  angle  (7  at  C 
and  D. 

With  0  as  a  centre,  and  with  the  same  radius,  describe 
the  indefinite  arc  AE. 

With  ^  as  a  centre,  and  with  a  radius  equal  to  the  chord 
CD,  describe  an  arc  intersecting  the  arc  AE  at  B]  and 
draw  OB. 

Then,  /.AOB=  ZO'. 

For  by  construction,  the  chords  of  the  arcs  AB  and  CD 
are  equal. 

Whence,  arc  AB  =  arc  CD.  (§  157.) 

Therefore,  Z  AOB  =  A  C.  (§  155.) 

EXERCISES. 

74.  Given  a  side  of  an  equilateral  triangle,  to  construct  the 
triangle. 

75.  Given  an  angle,  to  construct  its  complement. 

76.  Given  an  angle,  to  construct  its  supplement. 

77.  To  construct  an  angle  of  60°  (Ex.  74);  of  30°;  of  120°;  of  150°. 

78.  To  construct  an  angle  of  45°  ;  of  135°  ;  of  22^°  ;  of  67i°. 


THE   CIRCLE.  109 

Proposition  XXXIII.    Problem. 
210.    Given  two  angles  of  a  triangle^  to  find  the  third. 


yF 


jj/ 0 ^'- .D 

Let  A  and  B  be  two  angles  of  a  triangle. 
To  find  the  third  angle. 

Draw  the  indefinite  straight  line  CD;  and  at  any  point 
U  construct  Z  DEF  =  ZA,  and  Z  FEG  =ZB  (^  209). 

Then  CEG  is  the  required  angle. 

For  if  two  angles  of  a  triangle  are  given,  the  third  angle 
may  be  found  by  subtracting  their  sum  from  two  right 
angles.  (§  82.) 

Proposition  XXXIV.     Problem. 

211.  Through  a  given  point  without  a  given  straight  linSj 
to  draw  a  varallel  to  the  line. 


-D 


/E 


Let  C  be  the  given  point  without  the  straight  line  AB. 
■    To  draw  through  C  a  parallel  to  AB. 

Through  C  draw  any  line  EF,  meeting  AB  at  E. 
Construct  Z  FCD  =  Z  CEB  (§  209). 
Then  CD  is  parallel  to  AB. 

For  since  AB  and  CD  are  cut  by  EF,  making  the  corre- 
sponding angles  equal,  AB  and  CD  are  parallel.  (§  76.) 


110  PLANE   GEOMETRY.  —  BOOK  II. 

Peoposition  XXXV.     Problem. 

212.    Given  two  sides  and  the  included  angle  of  a  triangle^ 
to  construct  the  triangle. 


Let  m  and  n  be  the  given  sides,  and  A'  their  included 
angle. 

To  construct  the  triangle. 

Draw  the  straight  line  AB  equal  to  m. 
Construct  Z  BAD  =  Z  A. 
On  AD  take  AC  =  n,  and  draw  BC. 
Then  ABC  is  the  required  triangle. 

Proposition  XXXVI.     Problem. 

213.    Given  a  side  and  two  adjacent  angles  of  a  triangle.^ 
to  construct  the  triangle. 


Let  m  be  the  given  side,  and  A'  and  B'  its  adjacent  angles. 

To  construct  the  triangle. 

Draw  the  straight  line  AB  equal  to  m. 

Construct  Z  BAD  =  Z  A',  Siud  Z  ABE  =  Z  B'. 

Let  AD  and  BE  intersect  at  C. 

Then  ABC  is  the  required  triangle. 


THE   CIRCLE.   '  111 

214.  ScH.  I.  If  a  side  and  any  two  angles  of  a  triangle 
are  given,  the  third  angle  may  be  found  by  §  210,  and  the 
triangle  may  then  be  constructed  as  in  §  213. 

215.  ScH.  II.  The  problem  is  impossible  when  the  sum 
of  the  given  angles  is  equal  to,  or  greater  than,  two  right 
angles.  (§  82.) 

Proposition  XXXVII.     Problem. 

216.  Given  the  three  sides  of  a  triangle,  to  construct  the 
triangle. 


--^G- 


P 


Let  m,  n,  and  j)  be  the  given  sides. 
To  construct  the  triangle. 

Draw  the  straight  line  AB  equal  to  m. 

With  ^  as  a  centre,  and  with  a  radius  equal  to  n,  describe 
an  arc ;  with  ^  as  a  centre,  and  with  a  radius  equal  to  p, 
describe  an  arc  intersecting  the  former  arc  at  C, 

Draw  AC  and  BC. 

Then  ABC  is  the  required  triangle. 

217.  ScH.  The  problem  is  impossible  when  one  of  the 
given  sides  is  equal  to,  or  greater  than,  the  sum  of  the 
other  two.  (§  61.) 

EXERCISES. 

79.  Given  one  of  the  equal  sides  and  one  of  the  equal  angles  of 
an  isosceles  triangle,  to  construct  the  triangle. 

80.  To  construct  a  right  triangle,  having  given  the  hypotenuse 
and  an  acute  angle. 

81.  Through  a  given  point  without  a  straight  line  to  draw  a  lino 
making  a  given  angle  with  that  line. 


112  PLANE   GEOMETRY.  — BOOK  II. 

Proposition  XXXVIII.     Problem. 

218.  Given  two  sides  of  a  triangle,  and  the  angle  opposite 
to  one  of  them,  to  construct  the  triangle. 

Let  m  and  n  be  the  given  sides,  and  let  A '  be  the  angle 
opposite  to  n. 

To  construct  the  triangle. 

Construct  Z  DAE  =  A  A' ,  and  on  AE  take  AB  =  m. 

With  ^  as  a  centre,  and  with  a  radius  equal  to  n,  de- 
scribe an  arc. 

Case  I.    When  A'  is  acute,  and  m'^  n. 

There  may  be  three  cases : 

First.     The  arc  may  intersect  AD  in  two  points,  Cx  and 


m 


Bvsiw  BCi  and  j^Cg. 

There  are  then  two  triangles,  ABCi  and  ABC^,  either  of 
which  answers  to  the  given  conditions. 
This  is  called  the  ambiguous  case. 
Second.     The  arc  may  be  tangent  to  AD. 
In  this  case  there  is  but  one  solution ;  a  right  triangle. 

(§  170.) 
Third.     The  arc  may  not  intersect  AD  at  all. 

In  this  case  the  problem  is  impossible. 

Case  II.    When  A'  is  acute,  and  m  =  n. 
In  this  case,  the  arc  intersects  AD  in  two  points,  one  of 
which  is  A. 

There  is  then  but  one  solution  ;  an  isosceles  triangle. 


■ 


OF  THE      "^      \ 

UNIVERSITY  ) 


THE   CIRCLE. 


Case  III.    When  A'  is  acute,  and  m  <^n. 


113 


m 


C,\  A 


In  this  case,  the  arc  intersects  AD  in  two  points,  Ci  and 

But  the  triangle  ABCx  does  not  answer  to  the  given  con- 
ditions, since  it  does  not  contain  the  angle  A'. 

There  is  then  but  one  solution  ;  the  triangle  ABC^. 

Case  IV.    When  A'  is  right  or  obtuse,  and  m  <  n. 
In  each  of  these  cases,  the  arc   intersects   AD  in   two 
points  on  opposite  sides  of  A. 
There  is  then  but  one  solution. 

219.  ScH.  If  A'  is  right  or  obtuse,  and  w  =  ?i  or  m  >  n, 
the  problem  is  impossible ;  for  the  side  opposite  the  right  or 
obtuse  angle  in  a  triangle  must  be  the  greatest  side  of  the 
triangle.  (§  97.) 

The  student  should  construct  the  triangle  corresponding 
to  each  of  the  cases  of  §  218. 


EXERCISES. 

82.  To  construct  a  right  triangle,  having  given    a  leg  and  the 
opposite  angle. 

83.  Given  the  base  and  the  vertical  angle  of  an  isosceles  triangle, 
to  construct  the  triangle. 

84.  Given  the  altitude  and  one  of  the  equal  angles  of  an  isosceles 
triangle,  to  construct  the  triangle. 

85.  Given  two  diagonals  of  a  parallelogram  and  their  included 
fl,ngle,  to  construct  the  parallelogram. 


114  PLANE   GEOMETRY.— BOOK  II. 

Proposition   XXXIX.     Problem. 

220.    Given  two  sides  and  the  included  angle  of  a  paral- 
lelogram, to  construct  the  parallelogram. 


DjL 


Let  m  and  n  be  the  given  sides,  and  A'  their  included 
angle. 

To  construct  the  parallelogram. 

Draw  the  straight  line  AB  equal  to  m. 

Construct  Z  BAIJ  =  Z  A',  and  on  AJiJ  take  AB  =  n. 

With  ^  as  a  centre,  and  with  a  radius  equal  to  n,  de- 
scribe an  arc ;  with  Z>  as  a  centre,  and  with  a  radius  equal 
to  m,  describe  an  arc  intersecting  the  former  arc  at  C. 

Draw  ^C  and  DC. 

Then  ABCD  is  the  required  parallelogram. 

For  since,  by  construction,  AB  =  CD  and  AD  =  BC, 
ABCD  is  a  parallelogram.  (§  108.) 

Proposition-   XL.     Problem. 
221.    To  inscribe  a  circle  in  a  given  triangle. 


Let  ABC  be  the  given  triangle. 


THE  CIRCLE.  115 

To  inscribe  a  circle  in  ABC. 

Draw  AD  and  BE  bisecting  the  angles  A  and  B,  respec- 
tively (§  207);  and  from  their  intersection  0,  draw  OM 
perpendicular  to  AB. 

With  0  as  a  centre,  and  with  OM  as  a  radius,  describe  a 
circle. 

This  circle  will  be  tangent  to  AB,  BC,  and  CA. 

For  0  is  equally  distant  from  AB,  BC,  and  CA.    (§  135.) 

Proposition  XLI.     Problem. 
222.    To  circumscribe  a  circle  about  a  given  triangle. 


Let  ABC  be  the  given  triangle. 

To  circumscribe  a  circle  about  ABC. 

Draw  DF  and  EG  perpendicular  to  AB  and  AC,  respec- 
tively, at  their  middle  points  (§  205). 

Let  DF  and  EG  intersect  at  0. 

With  0  as  a  centre,  and  OA  as  a  radius,  describe  a 
circumference. 

This  circumference  will  pass  through  the  points  A,  B, 
and  C. 

For  0  is  equally  distant  from  A,  B,  and  C.  (§  137.) 

223.  ScH.  The  above  construction  serves  to  describe  a 
circumference  through  three  given  points  not  in  the  same 
straight  line,  or  to  find  the  centre  of  a  given  circumference 
or  arc. 


116 


PLANE   GEOMETRY.— BOOK  II. 


Proposition  XLII.     Problem. 

224.    To  draw  a  tangent  to  a  circle  through  a  given  point 
on  the  circumference. 


Let  A  be  the  given  point  on  the  circumference  AD. 
To  draw  through  A  a  tangent  to  AD. 

Draw  the  radius  OA. 

Through  A  draw  BG  perpendicular  to  OA  (§  203). 

Then  BC  will  be  tangent  to  AD.  (§  169.) 


Proposition  XLIII.     Problem. 

225.    To  draw  a  tangent  to  a  circle  through  a  given  point 
without  the  circle. 


Let  A  be  the  given  point  without  the  circle  BC. 
To  draw  through  A  a  tangent  to  BC. 

Let  0  be  the  centre  of  the  circle  BC. 
On  OA  as  a  diameter,  describe  a  circumference,  cutting 
.the  given  circumference  at  B  and  C]  and  draw  AB  and  AC. 


THE   CIRCLE. 


117 


Then  either  AB  ot  AC  is  tangent  to  BC. 
For  draw  OB. 

Then  Z  ABO,  being  inscribed  in  a  semicircle,  is  a  right 

angle.  (§  196.) 

Therefore,  AB  is  tangent  to  BC.  (§  169.) 

Proposition  XLIV.     Problem. 

226.    Upon  a  given  straight  line,  to  describe  a  segment 
which  shall  contain  a  given  angle. 


Let  AB  be  the  given  straight  line,  and  A'  the  given  angle. 
To   describe  a  circumference  AMBN,  such  that  every 
angle  inscribed  in  the  segment  AMB  shall  be  equal  to  A'. 

Construct  ZBAC  =  /.A'. 

Draw  DE  perpendicular  to  AB  at  its  middle  point  (§  205). 

Draw  AF  perpendicular  to  AC,  meeting  DE  at  0. 

With  0  as  a  centre,  and  OA  as  a  radius,  describe  the 
circumference  AMBN. 

Then  AMB  will  be  the  required  segment. 

For,  let  AGB  be  any  angle  inscribed  in  AMB. 

Then,  Z  AGB  is  measured  by  \  arc  ANB.  (§  193.) 

But  ^C  is  tangent  to  the  circle  AMB.  (§  169.) 

Whence,  Z  BAC  is  measured  by  \  arc  ANB.         (§  197.) 

Then,  Z  AGB  =  Z  BAC  =  Z  A\ 

Hence,  every  angle  inscribed  in  the  segment  AMB  is 
equal  to  A'.  (§  195.) 


118  PLANE   GEOMETRY.  —  BOOK  II. 


EXERCISES. 

86.  Given  the  middle  point  of  a  chord  of  a  circle,  to  construct 
the  chord. 

87.  To  circumscribe  a  circle  about  a  given  rectangle. 

88.  To  draw  a  line  tangent  to  a  given  circle  and  parallel  to  a 
given  straight  line. 

89.  To  draw  a  line  tangent  to  a  given  circle  and  perpendicular 
to  a  given  straight  line. 

90.  Through  a  given  point  to  draw  a  straight  line  forming  an 
isosceles  triangle  with  two  given  intersecting  lines. 

91.  Given  the  base,  an  adjacent  angle,  and  the  altitude  of  a  tri- 
angle, to  construct  the  triangle. 

92.  Given  the  base,  an  adjacent  side,  and  the  altitude  of  a  tri- 
angle, to  construct  the  triangle. 

93.  To  construct  a  rhombus,  having  given  its  base  and  altitude. 

94.  Given  the  altitude  and  the  sides  including  the  vertical  angle 
of  a  triangle,  to  construct  the  triangle. 

95.  Given  the  altitude  of  a  triangle,  and  the  angles  at  the  ex- 
tremities of  the  base,  to  construct  the  triangle. 

96.  To  construct  an  isosceles  triangle,  having  given  the  base  and 
the  radius  of  the  circumscribed  circle. 

97.  To  construct  a  square,  having  given  its  diagonal.     (§  196.) 

98.  To  construct  a  right  triangle,  having  given  the  hypotenuse 
and  the  length  of  the  perpendicular  drawn  to  it  from  the  vertex  of 
the  right  angle. 

99.  To  construct  a  right  triangle,  having  given  the  hypotenuse 
and  a  leg. 

100.  Given  the  base  of  a  triangle  and  the  perpendiculars  from 
its  extremities  to  the  other  sides,  to  construct  the  triangle. 

101.  To  describe  a  circle  of  given  radius  tangent  to  two  given 
intersecting  lines. 

102.  To  describe  a  circle  tangent  to  a  given  straight  line,  having 
its  centre  at  a  given  point. 

103.  Tp  construct  a  circle  having  its  centre  in  a  given  line,  and 
passing  through  two  given  points  without  the  line. 

104.  In  a  given  straight  line  to  find  a  point  equally  distant  from 
two  given  straight  lines. 


THE   CIRCLE.  119 

105.  Given  a  side  and  the  diagonals  of  a  parallelogram,  to  con- 
struct the  parallelogram. 

106.  Through  a  given  point  within  a  circle  to  draw  a  chord  equal 
to  a  given  chord.'     (§  165.) 

107.  Through  a  given  point  to  describe  a  circle  tangent  to  a  given 
straight  line  at  a  given  point. 

108.  Through  a  given  point  to  describe  a  circle  of  given  radius, 
tangent  to  a  given  straight  line. 

109.  To  describe  a  circle  of  given  radius  tangent  to  two  given 
circles. 

110.  To  describe  a  circle  tangent  to  two  given  parallels,  and 
passing  through  a  given  point. 

111.  To  describe  a  circle  of  given  radius,  tangent  to  a  given  line 
and  a  given  circle. 

112.  To  construct  a  parallelogram,  having  given  a  side,  an  angle, 
and  the  diagonal  drawn  from  the  vertex  of  the  angle. 

113.  In  a  given  triangle  to  inscribe  a  rhombus,  having  one  of  its 
angles  coincident  with  an  angle  of  the  triangle. 

114.  To  describe  a  circle  touching  two  given  straight  lines,  one 
of  them  at  a  given  point. 

115.  In  a  given  sector  to  inscribe  a  circle. 

116.  In  a  given  right  triangle  to  inscribe  a  square,  having  one  of 
its  angles  coincident  with  the  right  angle  of  the  triangle. 

117.  To  inscribe  a  square  in  a  given  rhombus. 

118.  To  draw  a  common  tangent  to  two  given  circles. 

119.  Given  the  base,  the  altitude,  and  the  vertical  angle  of  a  tri- 
angle, to  construct  the  triangle.     (§  226.) 

120.  Given  the  base  of  a  triangle,   its  vertical  angle,  and  the 
median  drawn  to  the  base,  to  construct  the  triangle. 

121.  To  construct  a  triangle,  having  given  the  middle  points  of 
its  sides.     (§130.) 

122.  Through  a  vertex  of  a  triangle  to  draw  a  straight  line  equally 
distant  from  the  other  vertices. 

123.  Given  two  sides  of  a  triangle,  and  the  median  drawn  to 
the  third  side,  to  construct  the  triangle. 

124.  Given  the  base,  the  altitude,  and  the  diameter  of  the  cir- 
cumscribed circle  of  a  triangle,  to  construct  the  triangle. 


Note.     For  additional  exercises  on  Book  II.,  see  p.  224. 


BOOK  III 


THEORY    OF   PROPORTION.  —  SIMILAR 
POLYGONS. 


DEFINITIONS. 

227.  A  Proportion  is  a  statement  that  two  ratios  are 
equal. 

The  statement  that  the  ratio  of  a  to  J  is  equal  to  the  ratio 
of  c  to  d,  may  be  written  in  either  of  the  forms 

a  :  0  =  c  :  a,  or  -  =  -. 
b       d 

228.  The  first  and  fourth  terms  of  a  proportion  are 
called  the  extremes,  and  the  second  and  third  the  means. 

The  first  and  third  terms  are  called  the  antecedents,  and 
the  second  and  fourth  the  consequents. 

Thus,  in  the  proportion  a  :  b  =  c  :  d,  a  and  d  are  the  ex- 
tremes, b  and  c  the  means,  a  and  c  the  antecedents,  and  b 
and  d  the  consequents. 

229.  If  the  means  of  a  proportion  are  equal,  either  mean 
is  called  a  mean  2>^oportional  between  the  first  and  last 
terms,  and  the  last  term  is  called  a  third  proportional  to 
the  first  and  second  terms. 

Thus,  in  the  proportion  a:b  =  b  :c,b  \^  a  mean  propor- 
tional between  a  and  c,  and  c  a  third  proportional  to  a  and  b. 

230.  A  fourth  proportional  to  three  quantities  is  the 
fourth  term  of  a  proportion,  whose  first  three  terms  are 
the  three  quantities  taken  in  their  order. 

Thus,  in  the  proportion  a-.b  =c'.d,  d  is  a  fourth  propor- 
tional to  a,  5,  and  c. 

120 


THEORY   OF  PROPORTION.  121 

Proposition  I.     Theorem. 

231.  In  any  proportion,  the  product  of  the  extremes  is 
equal  to  the  product  of  the  means. 

Let  the  proportion  \)Q  a:h  =  c:d. 
To  prove  ad  =  be. 

By  §227,  ^  =  '-. 

Multiplying  both  members  of  the  equation  by  bd,  we  have 
ad  =  be. 

232.  Cor.  The  mean  proportional  between  two  quantities 
is  equal  to  the  square  root  of  their  product. 

Let  the  proportion  hQ  a:b  =b  :  c.    '  (1) 

To  prove  b  =  s/ac. 

We  have  from  (1),         b""  =  ac.  (§  231.) 

Whence,  b  =  -Vac. 

Proposition  II.     Theorem. 

233.  (Converse  of  Prop.  I.)  If  the  product  of  two  quan- 
tities is  equal  to  the  p7'oduct  of  two  others,  one  pair  may  be 
made  the  extremes,  and  the  other  pair  the  means,  of  a 
proportion. 

Let  -  ad  =  be. 

To  prove  a-.b  =  c:d. 

Dividing  both  members  of  the  given  equation  by  bdj 
ad  _  bo 
bd''  bd' 
a       c 

b  =  d- 

That  is,  a  :  b  =  c  :  d. 

In  like  manner,  a  :  c  =  b  :  d,  ^ 

b  :  a  =  d :  c,  etc.  ^ 


122  PLANE   GEOMETRY.  — BOOK  III. 

Proposition  III.     Theorem. 

234.  In  any  proportion,  the  terms  are  in  proportion  hy 
Alternation  ;  that  is,  the  first  term  is  to  the  third  as  the 
second  term  is  to  the  fourth. 

Let  the  proportion  hQ  a\h  =  c-.d.  (1) 

To  prove  a:  c  =  b:d. 

We  have  from  (1),         ad  =  he.  (§  231.) 

Whence,  a:c  =  b\d.  (§  233.) 

Proposition   IV.     Theorem. 

235.  In  any  proportion,  the  terms  are  in  proportion  hy 
Inversion  ;  that  is,  the  second  term  is  to  the  first  as  the 
fourth  term  is  to  the  third. 

Let  the  proportion  hQ  a  :  h  =  c  v  d.  (1) 

To  prove  h  :  a  =  d-.c. 

We  have  from  (1),        ad  =  he.  (§  231.) 

Whence,  h-.a^d-.c.  (  (§  233.) 

Proposition   V.     Theorem. 

236.  In  any  proportion,  the  terms  are  in  proportion  hy 
Composition  ;  that  is,  the  sum  of  the  first  two  terms  is  to 
the  first  term  as  the  sum  of  the  last  two  terms  is  to  the  third 
term. 

Let  the  proportion  h^  a\h  =  c.d.  (1) 

To  prove,  a  -\-h  \  a  =  c  -\-  d  \c. 

We  have  from  (1),         ad -=  he.  (§  231.) 

Adding  each  member  of  the  equation  to  ae, 
ac  -\-  ad  =  ac  -f-  he, 
or,  ^.(1+.^)  =  c  (a  +  /;).     ^ 

Whence,  a  -^  h  :  a  =  e  -{-  d:  c.         \  (§  233.) 

In  like  manner,     a  -{-  h  :  h  =  c  -\-  d  :  d. 


THEORY  OF  PROPORTION.  123 

Proposition  VI.     Theorem. 

237.  Tn  any  proportion,  the  terms  are  in  proportion  hy 
Division  ;  that  is,  the  difference  of  the  first  two  terms  is  to 
the  first  term  as  the  difference  of  the  last  two  terms  is  to  the 
third  term. 

Let  the  proportion  bea:i  =  c:fZ,  (1) 

in  which  a  is  greater  than  h,  and  c  greater  than  d. 

To  prove  a  —  h:a  =  c~  d-.c. 

We  have  from  (1),         ad  =  he.  (§  231.) 

Subtracting  both  members  of  the  equation  from  ac, 
ac  —  ad  =  ac  —  be, 
or,  a  {c  —  d)  ^=  c  (a  —  b). 

Whence,  a —  b  :  a  =  c  —  d:c.  (§  233.) 

In  like  manner,     a  —  b -.b  =  c  —  d:d. 

PijoposiTioN  VII.     Theorem. 

238.  In  any  proportion,  the  terms  are  in  jyroportion  by 
Composition  and  Division  ;  that  is,  the  sum  of  the  first 
two  terms  is  to  their  difference  as  the  siim  of  the  last  two 
terms  is  to  their  difference. 

Let  the  proportion  he  U-.h  =  c:  d,  (1) 

in  which  a  is  greater  than  b,  and  c  greater  than  d. 
To  prove         a-\-b:  a  —  b  =  c-\-d:  c  —  d. 

We  have  from  ( 1 ),  ^-+^  =  ^±^ ,  (§  236.) 

a  c 

and  a^-b^^c^-d.  (§237.) 

a  c 

Dividing  the  first  equation  by  the  second,  we  have 

a  -^b  _  c  -\-  d 

a  —  b       c  —  d 

That  is,  a-{-b:a  —  b  =  c-\-d:c  —  d. 


124  PLANE   GEOMETRY. —BOOK  III. 


Proposition  VIII.     Theohem. 

239-  In  a  series  of  equal  ratios,  any  antecedent  is  to  its 
consequent  as  the  sum  of  all  the  aiitecedents  is  to  the  sum  of 
all  the  consequents.  . 

Let  a-.b  =^  c:  d  =^  e:f. 

To  prove        a:h  =  a-\-c-\-e:b-^d-\-f. 

Let  r  denote  the  value  of  each  of  the  given  ratios. 

Then,  <L  =  t      t  =  r. 

^        d      f 
Whence,  a  =  br,  c  =  dr,  e  =  fr. 

Then,  a -\-  c -\-  e  =  br  -\-  dr -\-  fr 

=  r(bJ^d-\-f). 

Whence,  ^il~\    l='^  =  T" 

That  is,  a:b  =  a-\-c-{-e:b-{-d-^f. 


Proposition   IX.     Theorem. 

240.  Equimultiples  of  two  quantities   are   in   the  same 
ratio  as  the  quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 
To  prove  ma  :  mb  =  a\b. 

We  have,  »m  _  «  ;     c^ 

mb       b  ^ 

Whence,  ma  :  mb  =  a\b. 

Proposition  X.     Theorem. 

241.  In  any  number  of  proportions,  the  products  of  the 
corresponding  terms  are  in  proportion. 

Let,  a:b  =  c:  d, 

and  e  :  f  =  g  :  h. 

To  prove  ae-.bf  =  eg  :  dh. 


THEORY  OF  PROPORTION.  125 


We  have,  ^  =-,and^  =  ^. 

b       d'        f      h 

Multiplying  these  equations,  we  obtain 
b   ^  f~  d  ^V 

'  bf       dh 

That  is,  aeibf  =  eg:  dh. 


Proposition  XI.     Theorem. 

^     242.   In  any  proportion,  like  powers  or  like  roots  of  the 
terms  are  in  proportion. 

Let  the  proportion  he  a  :  b  =  c  :  d.  (1) 

To  prove  a"  :  ^t"  =  c«  :  d". 

We  have  from  ( 1 ),  -  =  -  . 

b       d 

Eaising  both  members  to  the  nth  power, 
a^_  c^ 

That  is,  a«:  J«  =  c«:<Z". 

In  like  manner,    -^a  :  V^  =  Vc:  -y/d. 

Note.  The  ratio  of  two  magnitudes  of  tlie  same  kind  is  equal 
to  the  ratio  of  their  numerical  measures  when  referred  to  a  common 
unit  (§  183). 

Hence,  in  any  proportion  involving  the  ratio  of  two  magnitudes 
of  the  same  Kind,  we  may  regard  the  ratio  of  the  magnitudes  as 
replaced  by  the  ratio  of  their  numerical  measures  when  referred 
to  a  common  unit. 

Thus,  let  AB,  CD,  EF,  and  GH  be  four  lines  such  that 
AB:Cn=EF:GH. 

Then,  AB  x  GH  =  CD  x  EF.  (§  231. ) 

This  means  that  the  product  of  the  numerical  measures  of  AB  and 
GH  is  equal  to  the  product  of  the  numerical  measures  of  CD  and  EF. 

An  interpretation  of  this  nature  must  be  given  to  all  applications 
of  §§  231,  232,  241,  and  242. 


126  PLANE   GEOMETRY.  —  BOOK   III. 


EXERCISES. 

1.  Find  a  fourth  proportional  to  65,  80,  and  91. 

2.  Find  a  mean  proportional  between  28  and  63. 

3.  Find  a  third  proportional  to  |  and  |. 

4.  What  is  the  second  term  of  a  proportion  whose  first,  third,  and 
fourth  terms  are  45,  160,  and  224  ? 


PROPOETIONAL  LINES. 

Proposition  XII.     Theorem. 

243.  If  a  series  of  parallels,  cutting  ttvo  straight  lines, 
intercept  equal  distances  on  one  of  these  lines,  they  also  inter- 
cept equal  distances  on  the  other. 


Let  the  lines  AB  and  AB'  be  cut  by  the  parallels  CC, 
DD\  EE',  and  FF',  so  that 

CD  =  DE  =  EF. 
To  prove  CD'  =  D'E'  =  E'F'. 

In  the  trapezoid  CC'E'E,  DD'  is  parallel  to  the  bases, 
and  bisects  CE -,  it  therefore  bisects  C'E'.  (§  133.) 

That  is,  C'D'  =  D'E'.  (1) 

In  like  manner,  in  the  trapezoid  DD'F'F,  EE'  is  par- 
allel to  the  bases,  and  bisects  DF. 

Whence,  D'F/  =  E'F\  (2) 

From  (1 )  and  (2),  we  have 

CD'  =  D'E'  =  E'F'. .  ^        l2>L.>w»^ 


PROPOKTIOJ^AL  LINES.  127 

244.  Def.  Two  straight  lines  are  said  to  be  divided 
proportionally  when  their  corresponding  segments  are  in 
the  same  ratio  as  the  lines  themselves. 

A           E  B 

Thus,  the  lines  AB  and  CD  are  di-    ' J J 

vided  proportionally  if  G       F  D 

AE^BE^AB 
CF      DF      CD' 

Proposition  XIII.    Theorem. 

245.  A  parallel  to  one  side  of  a  triangle  divides  the  other 
two  sides  proportionally. 

Case  I.  When  the  segments  of  each  side  are  commen- 
surable. 

A 

f/\ 


Let  DF  be  parallel  to  the  side  BC  of  the  triangle  ABC. 
Let  AD  and  DB  be  commensurable. 

rp  AD      AF 

lo  prove  = . 

^  DB      FC 

Let  AF  be  a  common  measure  of  AD  and  DB ;  and  sup- 
pose it  to  be  contained  4  times  in  AD,  and  3  times  in  DB. 

Then,  ^  =  1  (1) 

DB      3  ^    ^ 

Drawing  parallels  to  ^C  through  the  several  points  of 

division,  AF  will  be  divided  into  4  parts,  and  FC  into  3 

parts,  all  of  which  parts  will  be  equal.  (§  243.) 

Whence,    .  ||  =  |.  (2) 

From  (1)  and  (2),       g  =  |f.  (Ax.  1.) 


128  PLANE  GEOMETRY.— BOOK  III. 

Case  II.    When  the  segments  of  each  side  are  incommen- 
surable. 


£ 


Let  DE  be  parallel  to  the  side  BC  oi  the  triangle  ABC. 
Let  AD  and  DB  be  incommensurable. 
'  AD      AE 

To  prove  m^EC' 

Let  AD  be  divided  into  any  number  of  equal  parts,  and 
let  one  of  these  parts  be  applied  to  DB  as  a  measure. 

Since  AD  and  DB  are  incommensurable,  a  certain  num- 
ber of  the  parts  will  extend  from  D  to  B',  leaving  a 
remainder  BB'  less  than  one  of  the  parts. 

Draw  B'C  parallel  to  ^C. 

Then,  m  =  ^''  (§245,  Case  I.) 

Now  let  the  number  of  subdivisions  of  AD  be  indefinitely 
increased. 

Then  the  length  of  each  part  will  be  indefinitely  dimin- 
ished, and  the  remainder  BB'  will  approach  the  limit  0. 

Then,  — — -  will  approach  the  limit  — , 

DB'  ^^  DB' 

A  TT  A  TP 

and  — —  will  approach  the  limit  -— . 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 

Whence,  AD^AE 

'  DB      EC 

246.   Cob.     The  result  of  §  245  may  be  written 

AD:DB  =  AE:EC.  (1) 

Then,         AD  +  DB  :  AD  =  AE  -{-  EC :  AE.        (§  236.) 


PROPORTIONAL   LINES.  129 

That  is,                 AB-.AD^AC:  AE.  (2) 

In  like  manner,    AB  :  DB  =  AC :  EC.  '  (3) 

From  (2),              AB'.AC  =  AD;  AE.  (§  234.) 

Also,  from  (3),     AB :  AC  =  DB  :  EC. 

^,,       .                     AB      AD      DB  /A\. 

therefore,               _  =  __  =  __.  (4) 

247.    ScH.     The  proportions  (1),  (2),  (3),  and  (4),  of 
§  246,  are  all  included  in  the  general  statement  of  §  245. 


Proposition  XIV.     Theorem. 

248.  (Converse  of  Prop.  XIII.)  A  line  which  divides 
two  sides  of  a  triangle  proportionally  is  parallel  to  the 
third  side. 


In  the  triangle  ABC,  let  DE  be  drawn  so  that 

AB  ^AC 

AD      AE' 
To  prove  DE  parallel  to  BC. 

Let  DF  be  drawn  parallel  to  BC. 

But  by  hypothesis,       -^  =  -^ . 

Then,  4^  =  4^- 

'  AE      AF 

Whence,  AE  =  AF. 

Then  DF  must  coincide  with  DE.  (Ax.  6.) 

Therefore,  DE  is  parallel  to  ^a 


130  PLANE   GEOMETKY.— BOOK   111. 


Proposition  XV.     Theorem. 

249.    In  any  trianyle,  the  bisector  of  an  angle  divides  the 
opposite  side  into  segments  proportional  to  the  adjacent  sides. 


Let  AD  be  the  bisector  of  the  angle  A  of   the  triangle 
ABC,  meeting  the  side  BC  at  D. 

To  prove  BB^AB 

^  ,  DC      AC 

Draw  B^  parallel  to  AD,  meeting  CA  produced  at  I^. 
Then  since  the  parallels  AD  and  B£J  are  cut  by  AB, 

Z  ABE  =  Z  BAD.  (§  72.) 

And  since  the  parallels  AD  and  BE  are  cut  by  CE, 

ZAEB=ZCAD.  (§74.) 

But  by  hypothesis, 

Z  BAD  =  Z  CAD. 
Whence,  Z  ABE  =  Z  AEB. 

Therefore,  AB  =  AE.  (§  91.) 

Now  since  AD  is  parallel,  to  the  side  BE  of  the  triangle 
BCE, 

DC      AC  ^  ^ 

Putting  for  AE  its  equal  AB,  we  have 

DB^AB 
DC      AC' 


PKOPORTIONAL  LINES.  131 

Proposition  XVI.    Theorem. 

250.  In  any  triangle,  the  bisector  of  an  exterior  angle 
divides  the  opposite  side  externally  into  segments  propor- 
tional to  the  adjacent  sides. 

Note.     The  theorem  does  not  hold  for  isosceles  triangles. 


Let  AD  be  the  bisector  of  the  exterior  angle  BAE  of  the 
triangle  ABC,  meeting  CB  produced  at  D. 

m  DB      AB 

To  prove  ^  =  2^* 

Draw  BF  parallel  to  AD,  meeting  AC  at  F. 

Then  since  the  parallels  AD  and  BF  are  cut  by  AB, 

Z  ABF  =  Z.  BAD.  (§  72.) 

And  since  the  parallels  AD  and  BF  are  cut  by  CE, 

Z  AFB  =  A  BAD.  (§  74.) 

But  by  hypothesis, 

Z  BAD  =  Z  FAD. 

Whence,  Z  ABF  =  Z  AFB. 

Therefore,  AB  =  AF.  (§  91.) 

Now  since  BF  is  parallel  to  the  side  AD  of  the  triangle 

ACD, 

—  =  ^.  (§245.) 

DC      AC  ^^        ^ 

Putting  for  AF  its  equal  AB,  we  have 

DB^AB 

DC~  AC' 


132  PLANE   GEOMETRY.— BOOK  III. 

251.  ScH.  The  segments  of  a  limited  straiglit  line  by  a 
point- are  understood  to  mean  the  distances  from  the  point 
to  the  extremities  of  the  line,  whether  the  point  is  in  the 
line  itself,  or  in  the  line  produced. 

EXERCISES. 

J  5.   The  sides  of  a  triangle  are  AB  =  8,  BC  =  Q,  and  C^  =  7;  find 

the  segments  into  which  BC  is  divided  by  the  bisector  of  the  angle  A. 

6.   The  sides  of  a  triangle  are  AB  --- 5,  BC  =  7,  and  CA^8;  find 

.'•y  the  segments  into  which  BC  is  divided  by  the  bisector  of  the  exterior 
angle  at  ^. . 

SIMILAR  POLYGONS. 

252.  Def.     Two  polygons  are  said  to  be  similar  when 
;  j    they   are   mutually   equiangular    (§  122),    and    have    their 

homologous  sides  proportional  (§  123). 


C 


ED  E'  D' 

Thus,  the  polygons  ABCDE  and  A'B'C'D'E'  are  similar 
if 

Z.A=ZA',  ZB=ZB',  ZC  =  Zr,  etc., 

and,  AB__BC_^C^ 

A'B'      B'C       CD" 

253.  ScH.  I.    The  following  forms  of   the  definition  of 
§  252  are  given  for  convenience  of  reference : 

I.    In  similar  polygons,  the  homologous  angles  are  equal. 
II.    In  similar  ^polygons,  the   homologous   sides   are  pro- 
portional. 

254.  ScH.  II.    In  similar  triangles,  the  homologous  sides 
lie  opposite  the  equal  angles. 


SIMILAR  POLYGONS.  13i 


Proposition  XVII.     Theorem. 

255.    Two  triangles  are  similar  when  they  are  mutually 
equiangular.  "^       j         V 


In  the  triangles  ABC  and  A'B'C,  let 

ZA=ZA',  AB=/.B',  and  /.  C  =  Z.  C. 
To  prove  ABC  and  A'B'C  similar. 

Superpose  the  triangle  A'B'C  upon  ABC,  so  that  Z  A' 
shall  coincide  with  Z  A  ;  the  side  B'C  falling  at  DE. 
Since  Z  ADE  =  ZB,DE  is  parallel  to  BC.  (§  76.) 

Therefore,  ^  =  ^.  CS  245  ) 

That  is,  A^  =  A^,  n\ 

In  like  manner,  by  superposing  A'B'C  upon  ^5C  so 
that  Z  ^'  shall  coincide  with  Z  ^.  we  may  prove 

AB        BC  ^2) 


From  (1)  and  (2), 


A'B'      B'C 
AB        AC        BC 


A'B'      A'C      B'C 
Then  ABC  and  A'B'C  have  their  homologous  sides  pro- 
portional, and  are  similar.  (§  252.) 

256.  CoR.  I.    Two  triangles  are  similar  when  two  angles 
of  one  are  equal  respectively  to  two  angles  of  the  other. 

For  their  remaining  angles  are  equal  each  to  each.  (§  86.) 

257.  Cor.  II.     Two    right  triangles    are    similar   when 
^      an  acute  angle  of  one  is  equal  to  an  acute  angle  of  the  other. 


% 


134 


PLANE   GEOMETRY.  —  BOOK  III. 


258.    Cor.  III.    If  a  line  be  drawn  between  two  sides  of 
a  triangle  parallel  to  the  third  side,  the 
triangle  formed  is  similar  to  the  given 
triangle. 

Let  DE  be  parallel  to  the  side  BC 
of  the  triangle  ABC. 

To  prove  the  triangle  ADE  similar 
to  ABC. 

We  have,  Z  ADE  =ZB.  (§  74.) 

Whence,  the  triangle  ADE  is  similar  to  ABC.       (§  256.) 


Proposition  XVIII.     Theorem. 

259.     Two  triangles  are  similar  when  their  homologous 
sides  are  proportional. 


In  the  triangles  ABC  and  A'B'C,  let 

AB  ^  AG_  ^  BC_ 

A'B'      AC      B'C' 
To  prove  ABC  and  AB'C  similar. 
Take  AD  =  AB',  and  AE  =  A'C,  and  draw  DE, 
Then  from  the  given  proportion,  we  have 

AB^AC 
AD      AE' 

Whence,  DE  is  parallel  to  BC.  (§  248. 

Then. the  triangles  ABC  and  ADE  are  similar.     (§  258. 

Therefore,       ^  =  ^,orA^-  =  ^. 
'       AD      DE'       AB'      DE 


(§  253,  11.) 


SIMILAR  POLYGONS.  135 

But  by  hypothesis,     ^  =  ^. 

Whence,  DE^^B'C, 

Therefore,  AADE  =A  A'B'  C\  (§  69.) 

But  ADE  has  been  proved  similar  to  J^BC. 

Hence,  AB' C  is  similar  to  ABC. 

Note.  To  prove  that  two  polygons  in  general  are  similar,  it  must 
be  shown  that  tliey  are  mutually  equiangular,  and  have  their  homol- 
ogous sides  proportional  (§252);  but  in  the  case  of  two  triangles^ 
each  of  these  conditions  involves  the  other  (§§  255,  259),  so  that 
it  is  only  necessary  to  show  that  one  of  the  tests  of  similarity  is 
satisfied. 

Proposition  XIX.    Theorem. 

260.  Two  triangles  are  siviilar  when  they  have  an  angle 
of  one  equal  to  an  angle  of  the  other ^  and  the  sides  including 
these  angles  proportional. 


In  the  triangles  ABC  and  ^'^'C",'let 

AB        AC 


ZA=Z  A',  and 


A'B'      A'C 
To  prove  ABC  and  AB'C  similar. 

Superpose  the  triangle  A'B'C  upon  ABC,  so  that  /.  A' 
shall  coincide  with  Z  A  ;  the  side  B'C  falling  at  DE. 

Then  by  hypothesis,    ^  =  A2 , 
^     ^^  'AD      AE 

Whence,  DE  is  parallel  to  BC.  (§  248.) 

Then  the  triangles  ABC  and  ADE  are  similar.      (§  258.) 

That  is,  the  triangles  ABC  and  A' B'C  are  similar. 


136 


PLANE   GEOMETKY.— BOOK  III. 


Proposition  XX.     Theorem. 

261.     Two   triangles    are   similar  when   their  sides   are 
parallel  each  to  each,  or  perpendicular  each  to  each. 


Fig.  1. 


Fig.  2. 


Fig.  .?. 


Let  the  sides  AB,  AC,  and  BC  of  the  triangle  ABC  be 
parallel  respectively  to  the  sides  A'B',  A'C,^  and  B'C  of 
the  triangle  A'B'C  (Fig.  2),  and  perpendicular  respectively 
to  the  sides  A'B',  A'C,  and  B'C  of  the  triangle  A'B'C 
(Fig.  3). 

To  prove  the  triangles  similar. 

Since  the  sides  of  the  angles  A  and  A'  are  parallel  each 
to  each,  or  perpendicular  each  to  each,  the  angles  are  either 
equal  or  supplementary.  (§§  79,  80,  81.) 

In  like  manner,  B  and  B\  and  C  and  C,  are  either  equal 
or  supplementary. 

We  may  then  make  five  hypotheses  with  regard  to  the 
angles  of  the  triangles,  B  denoting  a  right  angle  : 

1.  A-\-A'  =  2B,B-\-B'  =  2B,  (7+  C  =  2B. 

2.  A+A' =  2R,B^B' =  2R,  C    =C'. 
Z.   A^A'  =  2R,          B  =B'',     C-\-C'  =  2B. 

4.  A  =A,    B-{-B'  =  2E,  C-\-  C  =  2E. 

5.  A  =  A',  and     B  =B';  then,  C  =  C      (§  86.) 

But  the  first  four  hypotheses  are  impossible ;  for,  in 
either  case,  the  sum  of  the  angles  of  the  two  triangles 
would  exceed  four  right  angles.  (§  82.) 

We  can  then  have  only  A  =  A\  B  =  B',  and  C  =  C. 

Therefore,  the  triangles  are  similar.  (§  255.) 


SIMILAR  POLYGONS. 


137 


262.  ScH.  1.  In  similar  triangles  whose  sides  are  par- 
allel each  to  each,  the  parallel  sides  are  homologous. 

2.  In  similar  triangles  whose  sides  are  perpendicxdar  each 
to  each,  the  perpendicular  sides  are  homologous. 


Proposition  XXI.     Theorem. 

263.    The  homologous  altitudes  of  two  similar  triangle 
are  in  the  same  ratio  as  any  two  homologous  sides. 


B'  '    D 


Let  AD  and  A'l/  be  homologous  altitudes  of  the  similar 
triangles  ABC  and  A'B'C. 

T       oe         AD  ^  AB  ^  AC  _    BC 

"^^^  ^         A'D'      AB'      AC       B'C  ' 
Since  the  triangles  ABC  and  AB'C  are  similar, 

AB=AB'.  (§253,1.) 

Then  the  right  triangles  ABD  and  AB'V  are  similar. 

(§  257.) 

Whence,  ^'  =  ^r  (§253,11.) 

But  since  the  triangles  ABC  and  AB'C  are  similar, 

AB         AC        BC  (§253,11.) 


Whence, 


AB'      AC      B'C 
AD        AB        AC         BC 


AD'      AB'      AC      B'C 


264.    ScH.    In  two  similar  triangles,  any  two  homologous 
lines  are  in  the  same  ratio  as  any  two  homologous  sides. 


^ 


138  PLANE   GEOMETRY.— BOOK  III. 


Proposition  XXII.     Theorem. 

265.  If  two  ^parallels  are  cut  by  three  or  more  straight 
lines  passing  through  a  common  point,  the  corresponding 
segments  are  proportional. 


Let  the  parallels  AD  and  A'D'  be  cut  by  the  lines  OA, 
OB,  OC,  and  OD,  in  the  points  A,  B,  C,  D,  and  A',  B' , 
C,  D',  respectively. 

^   AB         BC  CD 


To  prove 


A'B'      B'C        CD' 


The  triangles  OAB  and  OA'B'  are  similar.  (§  258.) 

In  like  manner,  the  triangles  OBC  and  OB' C  are  similar. 

Whence,  ^  = -^^  =  ^ .  (2) 

'  OB'       B'C        OC  ^   ^ 

Again,  the  triangles  OCD  and  OCD'  are  similar. 

Whence,  ^  =  _^.  (3) 

'  OC       CD'  ^   ^ 

From  (1),  (2),  and  (3),  we  have 

AB         BC  CD 


A'B'      B'C        CD' 


EXERCISES. 


,        7.   The  sides  of  a  triangle  are  5,  7,  and  9.     The  shortest  side  of  a 
similar  triangle  is  14.     What  are  the  other  two  sides  ? 

8.   Two  isosceles  triangles  are  similar  when  their  vertical  angles 
are  equal. 


SIMILAR  POLYGONS.  139 


Proposition  XXIII.     Theorem. 

266.  Two  polygons  are  similar  when  they  are  composed  of 
the  same  number  of  triangles^  similar  each  to  each,  and 
similarly  placed. 


In  the  polygons  A  —  E  and  A'  —  E',  let  the  triangle  ABE 

be  similar  to  A'B'E',  BCE  to  B'C'E',  and  CDE  to  C'B'E\ 

To  prove  A—E  and  A—  E'  similar.  • 

Since  the  triangles  ABE  and  AB'E'  are  similar, 

/.A=AA!.  (§253,1.) 

Also,  A  ABE  =  A  AB'E',  (1) 

And  since  the  triangles  BCE  and  B'C'E'  are  similar, 

ZEBC  =  ZE'B'C',  (2) 

Adding  (1)  and  (2), 

ZABC  =  AAB'C'. 
In  like  manner,      ZBCD=Z  B'C'D',  etc. 
That  is,  A  —  E  and  A  —  E'  are  mutually  equiangular. 
Again,  since  ^^^  is  similar  to  AB'E',  SLudBCE  to  B'C'E', 
AB         BE          ,    BE  BC       ,.  o^q  tt  ^ 

AB'  =  WW^^''^WW  =  B^'^^  '''^  "'^ 
AB         BC 


Therefore, 

In  like  manner, 


AB'       B'C 
AB         BC  CD 


3'       B'C       CD'  ' 

That  is,  ^  —  ^  and  A—E'  have  their  homologous  sides 
proportional. 

Therefore,  A  —  E  and  A  —  E'  are  similar.  (§  252.) 


140  PLANE  GEOMETEY.  — BOOK  III. 


Proposition  XXIV.     Theorem. 

267.  (Converse  of  Prop.  XXIII.)  Two  similar  polygons 
may  be  decomposed  into  the  same  number  of  triangles,  similar 
each  to  each^  and  similarly  placed. 


Let  E  and  E'  be  homologous  vertices  of  the  similar  poly- 
gons A-E  and  A  -  E%  and  draw  EB,  EC,  E'B',  and  E'C. 
,  To  prove  the  triangle  ABE  similar  to  A'B'E',  BCE  to 
B'C'E',  and  CBE  to  C'B'E'. 

Since  tlie  polygons  are  similar, 

Z^=Z^',and^  =  Jf.  (1253.) 

Then,  the  triangles  ABE  and  AB'E'  are  similar.   (§  260.) 
Again,  since  the  polygons  are  similar, 

AABC  =  AAB'C',  (1) 

And  since  the  triangles  ABE  and  AB' E'  are  similar, 

ZABE  =  ZAB'E'.  (2) 

Subtracting  (2)  from  (1),  we  have 

AEBC=AE'B'C'. 
Also,  since  the  polygons  are  similar, 
AB   ^    BC 
A'B'       B'C 
And  since  the  triangles  ABE  and  A'B'E'  are  similar, 
AB  BE 

Whence, 


AB'       B'E' 

BC  _    BE 

BJC       B'E' 


SIMILAR  POLYGONS. 


141 


Then,  the  triangles  BCE  and  B'C'E'  are  similar.  (§  260.) 
In  like  manner,  we  may  prove  the  triangles  CDE  and 
C'D'E'  similar. 

Proposition  XXV.     Theorem.  . 

.  268-    The  'perimeters  of  two  similar  polygmis  are  in  the 
same  ratio  as  any  two  homologous  sides. 


Let  F  and  P'  denote  the  perimeters  of  the  similar  poly- 
gons A—E  and  A'  —  E',  respectively. 

P  ^  AB_  ^  BC         CD 
P' 


To  prove 


A'B'      B'C 


CD 


-,  etc. 


Since  the  polygons  are  similar,  we  have 

AB         BC  ^^_,etc.     (§253,11.) 


AB' 


B'C 


CD' 


,p,  AB  -{-  BC  -\-    CD  -f-etc.        AB  /roqqn 

^^-^      AW.^Wc^C'D'\.t.=AB''       (^'^^-^ 


Therefore, 


P 
P' 


AB 


BC 


AB'       B'C 


CD 
CD 


-,etc. 


EXERCISES. 

9.  The  base  and  altitude  of  a  triangle  are  5  ft.  10  in.  and  3  ft. 
(5  in.,  respectively.  If  the  homologous  base  of  a  similar  triangle  is 
V  ft.  6  in.,  find  its  homologous  altitude. 

10.  The  perimeters  of  two  similar  polygons  are  119  and  68;  if  a 
side  of  the  first  is  21,  what  is  the  homologous  side  of  the  second  ? 

11.  J^J5  is  the  hypotenuse  of  a  right  triangle  ABC  If  perpen- 
diculars be  drawn  to  AB  at  A  and  B,  meeting  AC  produced  at  D, 
and  BC  produced  at  i?,  prove  the  triangles  ACE  and  BCD  similar. 


142 


PLANE   GEOMETRY.— BOOK  III. 


Proposition  XXVI.    Theorem. 

269.    If  a  perpendicular  he  drawn  from  the  vertex  of  the 
right  angle  to  the  hypotenuse  of  a  right  triangle, 

I.    The  triangles  formed  are  similar  to  the  whole  tri- 
angle, and  to^  each  other. 

II.    The  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse. 

III.   Either  leg  is  a  mean  proportional  between  the  whole 
hypotenuse  and  the  adjacent  segment. 


Let  CD  be  perpendicular  to  the  hypotenuse  AB  of  the 
right  triangle  ABC. 

I.  To  prove  that  the  triangles  ACD  and  BCD  are  similar 
to  ABC,  and  to  each  other. 

In  the  right  triangles  ACD  and  ABC,  Z  A  is  common. 
Whence,  ACD  is  similar  to  ABC.  (§  257.) 

In  like  manner,  BCD  is  similar  to  ABC. 
Then  A  CD  and  B  CD  are  similar,  for  each  is  similar  to  AB  C. 

II.  To  prove        AD:  CD  =  CD:  BD. 

Since  the  triangles  ACD  and  BCD  are  similar, 

Z  AQD  =  ZB,Sind  ZA  =  Z  BCD.    (§  253,  I.) 

Then  AD,  the  side  of  ACD  opposite  Z  ACD,  is  homolo- 
gous to  CD,  the  side  of  BCD  opposite  Z  B;  and  CD,  the 
side  of  ACD  opposite  Z  A,  is  homologous  to  BD,  the  side 
of  BCD  opposite  Z  BCD.  (§  254.) 

Whence,  AD  :  CD  =  CD:  BD.  (§  253,  II.) 


SIMILAR  rOLYGONS.  143 

III.   To  prove       AB  :  AC  =  AC :  AD. 

Since  the  triangles  ABC  and  A  CD  are  similar, 

Z  ACB  =  Z  ADC,  and  Z  ^  =  Z  ACD.    (§  253, 1.) 

Then  AB,  the  side  of  yI^C  opposite  Z  ^C^,  is  homolo- 
gous to  AC,  the  side  of  ACD  opposite  ZADC;  and  AC, 
the  side  of  ^^C  opposite  Z  -S,  is  homologous  to  AD,  the 
side  of  ACD  opposite  Z  ^C2).  (§  254.) 

Whence,  AB :  AC  =  AC :  AD.  (§  253,  II.) 

In  like  manner,     AB:BC=BC:  BD. 

270.  CoR.  I.   The  tkree  proportions  of  §  269  give 

CD'  =  ADx  BD, 

AC''  =  ABx  AD, 

and  BC^  =-AB  X  BD.  (§  231.) 

Hence,  if  a  perpendicular  be  drawn  from  the  vertex  of  the 
right  angle  to  the  hypotenuse  of  a  right  trianglS, 

1.  The  square  of  the  perpendicular  is  equal  to  the  product 
of  the  segments  of  the  hypotenuse. 

II.  The  square  of  either  leg  is  equal  to  the  product  of  the 
whole  hypotenuse  and  the  adjacent  segment. 

•  As  stated  in  the  Note,  p.  125,  the  above  equations  mean 
simply  that  the  square  of  the  numerical  measure  of  CD  is 
equal  to  the  product  of  the  numerical  measures  of  AD  and 
BD,  etc. 

271.  CoR.  II.  Dividing  the  second  equation  of  §  270  by 
the  third,  we  obtain 

AC'^  ^AB  X  AD  ^AD 
BC'      ABXBD      BD 

That  is,  if  a  perpendicular  be  drawn  from  the  vertex  of 
the  right  angle  to  the  hypotenuse  of  a  right  triangle,  the 
squares  of  the  legs  are  proportional  to  the  adjacent  segments 
of  the  hypotenuse. 


144 


PLANE   GEOMETRY.— BOOK  III. 


272.  Cor.  III.  Since  an  angle  inscribed  in  a  semicircle 
is  a  right  angle  (§  196),  it  follows  that : 

If  a  perpendicular  be  drawn  from  any 
point  in  the  circumference  of  a  circle  to 
a  diameter, 

I.    The  perpendicular  is  a  mean  pro- 
portional between  the  segments  of  the  diameter. 

II.  The  chord  joining  the  point  to  either  extremity  of  the 
diameter  is  a  mean  proportional  between  the  whole  diameter 
and  the  adjacent  segment. 


Proposition^  XXVII.     Theorem. 

273.    In  any  right  triangle,  the  square  of  the  hypotenuse 
is  equal  to  the  sum  of  the  squares  of  the  legs.  .  .  _ 


Let  AB  be  the  hypotenuse  of  the  right  triangle  ABC. 
To  prove  AB^  =  AC''  +  BC\ 

Draw  CD  perpendicular  to  AB. 
Then,  Z^'  =  AB  x  AD, 

and  BC""  =  AB  X  BD.  (§  270,  II.) 

Adding,  Ic''  +  ~BC^  =  AB  x  {AD  +  BD) 

=  AB  xAB  =  AB\ 


AC\ 


274.    Cor.  I.    It  follows  from  §  273  that 
'AC^  =  AB^-BC\2^nd.BC''  =  AB'' 

That  is,  in  any  right  triangle,  the  square  of  either  leg  is 
equal  to  the  square  of  the  hypotenuse,  minus  the  square  of 
the  other  leg. 


'J- 


h' 


SIMILAR  POLYGONS. 


145 


275.    CoK.  II.   If  ^C  is  a  diagonal  of  the  square  ABCDy 
we  have 
AC'  =  AB'  +BC'  =  AB^  ^AB'  =  2AB\    ^ ^ 

Dividing  each  member  of  the  equation 
hy  AB\ 

AB'  AB 

Hence,  the  diagonal  of  a  square  is  incommensurable  with 
its  side  (§  181). 


X 


EXERCISES. 

12.  What  is  the  length  of  the  tangent  to  a  circle  whose  diameter 
is  16,  from  a  point  whose  distance  from  the  centre  is  17  ? 

13.  What  is  the  length  of  the  longest  straight  line  which  can  be 
drawn  on  a  floor  33  ft.  4  in.  long,  and  16  ft.  3  in.  wide  ? 

14.  A  ladder  32  ft.  6  in.  long  is  placed  so  that  it  just  reaches 
a  window  26  ft.  above  the  street;  and  when  turned  about  its  foot, 
just  reaches  a  window  16  ft.  6  in.  above  the  street  on  the  other  side. 
Find  the  width  of  the  street. 

15.  The  side  of  an  equilateral  triangle  is  5 ;  what  is  its  altitude  ? 

16.  Find  the  length  of  the  diagonal  of  a  square  whose  side  is 
1  ft.  3  in. 

276.    Definitions.    The   projection  of  a  point  •upon  a 

straight  line  of  indefinite  length,  is 
the  foot  of  the  perpendicular  let  fall 
from  the  point  to  the  line. 

Thus,  if  AA  be  drawn  perpendicular 
to  CD,  the  projection  of  the  point  A 
upon  the  line  CD  is  the  point  A\ 

The  projection   of  a  finite  straight 
line  upon  a  straight  line  of  indefinite  length,  is  that  por- 
tion of  the  second  line  included  between  the  projections  of 
the  extremities  of  the  first. 

Thus,  if  AA'  and  BB'  be  drawn  perpendicular  to  CD,  the 
projection  of  the  line  AB  upon  the  line  CD  is  A'B'. 


C  A 


B'  D 


'^ 


146 


PLANE   GEOMETRY.— BOOK  III. 


Proposition  XXVIII.     Theorem. 

277.  In  any  triangle,  the  square  of  the  side  opposite  an 
acute  angle  is  equal  to  the  sum  of  the  squares  of  the  other 
two  sides,  minus  twice  the  product  of  one  of  these  sides  and 
the  projection  of  the  other  side  upon  it. 


Fig.  2. 


Let  C  be  an  acute  angle  of  the  triangle  ABC,  and  draw 
AD  perpendicular  to  CB,  produced  if  necessary. 

Then  CD  is  the  projection  of  the  side  ^C  upon  BC  (^  276). 
To  prove  AB^  =  W  +  2^'  —  2BCx  CD. 

There  will  be  two  cases  according  as  D  falls  upon  CB 
(Fig.  1),  or  upon  CB  produced  (Fig.  2). 
In  Fig.  1,  BD  =  BC-CD. 

In  Fig.  2,  BD=CD-BC. 

Squaring  these  equations,  we  have  in  either  case 

BD"  =  BC''  +  CD"  -  2  BC  X  CD. 
Adding  AD  to  both  members, 

ad'  +  'BD^  =  BC''  +  Id"  j^C&-2BCx  CD. 


But  in  the  right  triangles  ABD  and  ACD, 


and 


ADi' 
AD" 


+  BD'  =  AB', 


+  CD' 


Whence,  AB^  =  BC"" -^  AC^ 


AC  . 
2BCXCD. 


(§  273.) 


I 


Ex.  17.   The  square  of  the  side  of  an  equilateral  triangle  is  equal 
to  four-thirds  the  square  of  the  altitude, 


SIMILAR  POLYGONS.  147 


Proposition  XXIX.    Theorem. 

278.  In  any  triangle  having  an  obtuse  angle,  the  square 
of  the  side  op2iosite  the  obtuse  angle  is  equal  to  the  sum  of 
the  squares  of  the  other  two  sides,  plus  twice  the  product  of 
one  of  these  sides  and  the  projection  of  the  other  side  upon  it. 


Let  C  be  an  obtuse  angle  of  the  triangle  ABC,  and  draw 
AD  perpendicular  to  BC  produced. 

Then  CD  is  the  projection  of  the  side  AC  upon  BC. 
To  prove  IB''  =  B(?  +  Jc'  +  25C  X  CD. 

We  have,  BD  =  BC  -\-  CD. 

Squaring  this  equation, 

BD"  ^  BC^  j^CD"  +  2BCX  CD. 
Adding  AD  to  both  members, 

A&  4-  ISD"  =  'BC'-  +  A&  -^-CD"  +  2BCX  CD. 
But  in  the  right  triangles  ABD  and  A  CD, 
AD"  +  BD"  =  AB\ 
and  AD"  -\-C&  =  AC\  (§  273.) 

Whence,   AB^  =  BC''  +  i^'  +  2  5(7  X  CD. 

EXERCISES. 

18.  The  length  of  the  tangent  to  a  circle,  whose  diameter  is  80,    -^ 
from  a  point  without  tlie  circumference,  is  42.    What  is  the  distance 

of  the  point  from  the  centre  ? 

19.  Find  the  length  of  the  common  tangent  to  two  circles  whose 
radii  are  11  and  18,  if  the  distance  between  their  centres  is  25. 


148  PLANE   GEOMETRY.  —  BOOK  III. 

Proposition  XXX.    Theorem. 

279.  In  any  triangle,  if  a  median  be  drawn  from  the 
vertex  to  the  base, 

I.  The  sum  of  the  squares  of  the  other  tico  sides  is  equal 
to  twice  the  square  of  half  the  base,  plus  twice  the  square  of 
the  median. 

II.  The  difference  of  the  squares  of  the  other  two  sides  is 
equal  to  twice  the  product  of  the  base  and  the  projection  of 
the  median  upon  the  base. 

C 


Let  DU  be  the  projection  of  the  median  CD  upon  the  base 
AB  of  the  triangle  ABC;  and  let  ^C  be  greater  than  BC. 
To  prove    I.   2o'  +  BC""  =  2  W  +  2  CD^. 
II.   AC^-BC^  =  2ABxBK 

Since  ^  falls  to  the  right  of  D,  Z  ABC  is  obtuse,  and 
ZBDC  is  acute. 
Hence,  in  the  triangles  ABC  and  BBC, 

AC''  =^A&  -^CB^  +  2ADXBE,         (§278.) 
and  BC''  =  B&  +  CBt  -  2  BD  X  BE.          (§  277.) 

Or  since  BB  =  AD, 

AC^  =  AB' -}- CB'' -{-  ABxBE,  (1) 

and  'W  =  A&  -}-  CD"  -  AB  x  BE.  (2) 

Adding  (1)  and  (2),  we  have 

AC''  +  BC'  ==2AB'  +  2CD\ 
Subtracting  (2)  from  (1), 

f2 


AC'-BC'  =  2ABxBE. 


SIMILAR  POLYGONS.  149 

Proposition  XXXI.     Theorem. 

280.  If  any  two  chords  he  drawn  through  a  fixed  point 
within  a  circle,  the  product  of  the  segments  of  one  chord  is 
equal  to  the  product  of  the  segments  of  the  other. 

>  ;  '^=^ 


Let  AB  and  A'B'  be  any  two  chords  passing  through  the 
fixed  point  F  within  the  circle  ABB'. 

To  prove  APxBP  =  A'P  XB'F. 

Draw  AA'  and  BB'. 

Then  in  the  triangles  AA'F  and  BB'F, 
ZAA'F=  ZB'BF, 
since  each  is  measured  by  ^  arc  AB '.  (§  193.) 

In  like  manner,     Z  AAF  =  Z  BB'F. 

Then  the  triangles  AA'F  and  BB'F  are  similar.    (§  256.) 

Whence,  AF  :A'F=B'F:  BF.   (§  §  253,  II.,  254.) 

Therefore,  AF  xBF  =  A'F  x  B'F.  (§  231.) 

281.    ScH.     The  proportion  in  §  280  may  be  written 
AF  ^  B;F  ^^  AF  ^  J_ 
A'F       BF '  ^^  A'F       BF  ' 


B'F 

If  two  magnitudes,  such  as  the  segments  of  a  chord  by  a 
point,  are  so  related  that  the  ratio  of  any  two  values  of  one 
is  equal  to  the  reciprocal  of  the  ratio  of  the  corresponding 
values  of  the  other,  they  are  said  to  be  reciprocally  pro- 
portional. 

Thus,  if  any  two  chords  he  drawn  through  a  fixed  point 
within  a  circle,  their  segments  are  reciprocally  proportional. 


150  PLANE  GEOMETRY.— BOOK  III. 


Proposition  XXXII.     Theorem. 

282.  If  through  a  fixed  point  without  a  circle  a  secant  and 
a  tangent  be  drawn,  the  product  of  the  whole  secant  and  its 
external  segment  is  equal  to  the  square  of  the  tangent. 


Let  AP  be  a  secant,  and  CP  a  tangent,  passing  through 
the  fixed  point  P  without  the  circle  ABC. 

To  prove  APxBP  =  CP''. 

Draw  ^(7  and  5  a 

Then  in  the  triangles  ACP  and  BCP,  Z  P  is  common. 

Also,  ZA=Z  BCP, 

since  each  is  measured  by  ^  arc  BC.  (§§  193,  197.) 

Then  the  triangles  ACP  and  BCP  are  similar.       (§  256.) 

Whence,  AP  :  CP=CP:  BP.     (§§  253,  II.,  254.) 

Therefore,  AP  X  BP  =  CP''.  (§  231.) 

283.  Cor.  I.  If  through  a  fixed  point  without  a  circle  a 
secant  and  a  tangent  be  drawn,  the  tangent  is  a  mean  pro- 
portional between  the  whole  secant  and  its  external  segment. 

284.  Cor.  II.     Let  AP  and  A'P  be  any  two  secants  of 
the  circle  ACB,  intersecting  without 
the   circumference ;    and   draw   CP 
tangent  to  the  circle  at  C. 

Then,    APxBP=  CP^ 
and  A'P  X  B'P  =  CP\  (§  282.) 

Hence,  AP  xBP  =  A'P  x  B'P. 


SIMILAR  POLYGONS.  151 

That  is,  if  any  two  secants  he  drawn  through  a  fixed  point" 
without  a  circle,  the  product  of  one  and  its  external  segment 
is  equal  to  the  product  of  the  other  and  its  external  segment. 

285.  Cor.  III.  If  any  two  secants  be  drawn  through  a 
fixed  pjoint  without  a  circle,  the  whole  secants  and  their  ex- 
ternal segments  are  reciprocally  proportional  (§  281). 

Proposition  XXXIII.     Theorem. 

286.  In  any  triangle,  the  product  of  any  two  sides  is  equal 
to  the  diameter  of  tlm  circumscribed  circle,  multiplied  hy  the 
perpendicular  drawn  to  the  third  side  from  the  vertex  of  the 
opposite  angle. 


Let  AD  be  the  diameter  of  the  circumscribed  circle  ACD 
of  the  triangle  ABC,  and  AE  the  perpendicular  from  A  to  BC. 
To  prove  AB  X  AC  =  AD  X  AE. 

.  Draw  BD',  then,  Z  ABD  is  a  right  angle.  (§  196.) 
Now  in  the  right  triangles  ABD  and  ACE, 
ZD=  Z.C, 

since  each  is  measured  by  ^  arc  AB.  (§  193.) 

Then  the  triangles  ABD  and  ACE  are  similar.  (§  257.) 

Whence,                AB  :  AD  =  AE :  AC.    (§ §  253,  II.,  254.) 

Therefore,           ABxAC=ADxAE.  (§  231.) 

287.  Cor.  In  any  triangle,  the  diameter  -of  the  circum- 
scribed circle  is  equal  to  the  product  of  any  two  sides  divided 
by  the  perpeiidicular  drawn  to  the  third  side  from  the  veHex 
of  the  opposite  angle. 


152  PLANE   GEOMETKY.  — BOOK   III. 


Proposition  XXXIV.    Theorem. 

288.  In  any  triangle,  the  product  of  any  two  sides  is 
equal  to  the  product  of  the  segments  of  the  third  side  formed 
by  the  bisector  of  the  opposite  angle,  p)lus  the  square  of  the 
bisector. 


In  the  triangle  ABC,  let  AD  bisect  the  angle  A. 
To  prove  AB  xAC  =  BD  xDC -^  AD\      . 

Circumscribe  a  circle  about  ABC. 

Produce  AD  to  meet  the  circumference  at  E,  and  draw 
CE. 
Then  in  the  triangles  ABD  and  ACE,  by  hypothesis, 

A  BAD  =  A  CAE. 
Also,  ZB=Z.E, 

since  each  is  measured  by  ^  arc  AC.  (§  193.) 

Then  the  trianglesyi-BD  and  ACE  are  similar.      (§  256.) 
Whence,  AB  :  AD  =  AE :  AC.   (§§  253,  II.,  254.) 

Therefore,  ABxAC  =  ADxAE  (§  231.) 

=  ADx  (DE  +  AD) 
=  ADxDE  +  AJ)\ 
But,  ADxDE  =  BDx DC.  (§  280.) 

Whence,  AB  xAC  =  BD  xDC  -\-  AD\ 


Ex.  20.   If  AD  is  the  perpendicular  from  A  to  the  side  BC  of  ^^ 
the  triangle  ABC,  prove  that 


SIMILAR  POLYGONS.  153 


EXERCISES. 

21.  If  the  length  of  the  common  chord  of  two  intersecting  circles 
is  16,  and  their  radii  are  10  and  17,  what  is  the  distance  between 
cheir  centres  ? 

22.  If  one  leg  of  a  right  triangle  is  double  the  other,  the  perpen- 
dicular from  the  vertex  of  the  right  angle  to  the  hypotenuse  divides 
it  into  segments  which  are  to  each  other  as  1  to  4. 

23.  If  two  parallels  to  the  side  BC  of  a.  triangle  ABC  meet  the 
sides  AB  and  -4C  in  D  and  F,  and  E  and  (?,  respectively,  prove 

BD^BF^BF 
CE       CO      EG  * 

24.  C  and  D  are  the  middle  points  of  a  chord  AB  and  its  sub- 
tended arc.  If  AD  =  12  and  CD  =  8,  what  is  the  diameter  of  the 
circle  ? 

25.  If  ^D  and  BE  are  the  perpendiculars  from  the  vertices  A 
and  B  of  the  triangle  ABC  to  the  opposite  sides,  prove  that 

AC  :  DC  =  BC :  EC: 

26.  If  D  is  the  middle  point  of  the  side  BC  of  the  triangle  ABC, 
right-angled  at  C,  prove  that  ZB^  —  AD^  =  3  CD\ 

27.  The  diameters  of  two  concentric  circles  are  14  arid  50  units, 
respectively.  Find  the  length  of  a  chord  of  the  greater  circle  which 
is  tangent  to  the  smaller. 

28.  The  length  of  a  tangent  to  a  circle  from  a  point  8  units  dis- 
tant from  the  nearest  point  of  the  circumference,  is  12  units.  What 
is  the  diameter  of  the  circle  ? 

29.  The  non-parallel  sides  AD  and  BC  oi  &,  trapezoid  ABCD 
intersect  at  O.  If  AB  =  15,  CD  =  24,  and  the  altitude  of  the  trape- 
zoid is  8,  what  is  the  altitude  of  the  triangle  OAB  ? 

30.  If  the  equal  sides  of  an  isosceles  right  triangle  are  each  18 
units  in  length,  what  is  the  length  of  the  median  drawn  from  the 
vertex  of  the  right  angle  ? 

31.  The  non-parallel  sides  of  a  trapezoid  are  each  53  units  in 
length,  and  one  of  the  parallel  sides  is  56  imits  longer  than  the 
other.     Find  the  altitude  of  the  trapezoid. 

32.  AB  is  a  chord  of  a  circle,  and  CE  is  any  chord  drawn 
through  the  middle  point  C  of  the  arc  AB,  cutting  the  chord  AB 
at  D.     Prove  that  ^  C  is  a  mean  proportional  between  CD  and  CE. 


154  PLANE   GEOMETRY.  —  BOOK   III. 

33.  Two  secants  are  drawn  to  a  circle  from  an  outside  point.  If 
their  external  segments  are  12  and  9,  while  the  internal  segment 
of  the  former  is  8,  what  is  the  internal  segment  of  the  latter  ? 

34.  If,  in  a  triangle  ABC,  ZC  =  120°,  prove  that 

AB^  =  BC^  +  AC^  +  ACx  BC. 

35.  BC  is  the  base  of  an  isosceles  triangle  ABC  inscribed  in  a 
circle.    If  a  chord  AD  be  drawn  cutting  BC  at  E,  prove  that 

ABiAB  =  AB:  AE. 

36.  Two  parallel  chords  on  opposite  sides  of  the  centre  of  a  circle 
are  48  units  and  14  units  long,  respectively,  and  the  distance  between 
their  middle  points  is  31  units.     What  is  the  diameter  of  the  circle  ? 

37.  ABC  is  a  triangle  inscribed  in  a  circle.  Another  circle  is 
drawn  tangent  to  the  first  externally  at  C,  and  AC  and  BC  are 
produced  to  meet  its  circumference  at  B  and  E.  Prove  that  the 
triangles  ABC  and  CBE  are  similar.     (§  197.) 

38.  ABC  and  A'BC  are  triangles  whose  vertices  A  and  A'  lie  in 
a  parallel  to  their  common  base  BC.  If  a  parallel  to  BC  cuts  AB 
and  AC  at  D  and  E,  and  A'B  and  A'C  at  D'  and  JS",  prove  that 

BE=B'E\ 

39.  A  line  parallel  to  the  bases  of  a  trapezoid,  passing  through 
the  intersection  of  the  diagonals,  and  termiijating  in  the  non- 
parallel  sides,  is  bisected  by  the  diagonals. 

40.  If  the  sides  of  a  triangle  ABC  are  ^J5  =  10,  BC  =  14,  and 
CA  =  16,  find  the  lengths  of  the  three  medians.    (§  279,  I.) 

41.  If  the  sides  of  a  triangle  are  AB  =  4,  AC  =  5,  and  BC  =^  Q, 
find  the  length  of  the  bisector  of  the  angle  A.    (§§  249,  288.) 

42.  The  tangents  to  two  intersecting  circles  from  any  point  in  their 
common  chord  produced  are  equal.    (§  282.) 

43.  If  two  circles  intersect,  their  common  chord  produced  bisects 
their  common  tangents. 

44.  AB  and  AC  are  the  two  tangents  to  a  circle  from  the  point 
A.     If  CD  be  drawn  perpendicular  to  the  radius   OB  at  D,  prove 

AB  :  OB  =  BD  :  CD. 

45.  ABC  is  a  triangle  inscribed  in  a  circle.  A  straight  line  AD 
is  drawn  from  A  to  any  point  of  BC,  and  a  chord  BE  is  drawn, 
making  Z  ABE  =  Z  ADC.     Prove  that 

AB  X  AC  ^  AD  X  AE. 


SIMILAR  POLYGONS.  155 

46.  The  radius  of  a  circle  is  22^  units.  Find  the  length  of  a 
chord  wliich  joins  the  points  of  contact  of  two  tangents,  each  30 
units  in  length,  drawn  to  the  circle  from  a  point  without  the 
circumference. 

47.  If,  in  a  right  triangle^ J. 7^ C,J.he  acute  angle  B  is  double  the 
acute  angle  A,  prove  that  AC^  =  Q  B&. 

48.  What  is  the  product  of  the  segments  of  any  chord  drawn 
through  a  point  9  units  from  the  centre  of  a  circle  whose  diameter  is 
24  units  ? 

49.  The  hypotenuse  of  a  right  triangle  is  5,  and  the  perpendicular 
to  it  from  the  opposite  vertex  is  2|.  Find  the  other  two  sides  of  the 
triangle,  and  the  segments  into  which  the  perpendicular  divides  the 
hypotenuse.     (§270.) 

50.  State  and  prove  the  converse  of  Prop.  XY. 

51.  State  and  prove  the  converse  of  Prop.  XVI. 

52.  If  D  is  the  middle  point  of  the  hypotenuse  AB  of  the  right 
triangle  ABC^  prove  that 

CD^=\  (AB^  +  BC^  +  CA^). 

53.  If  a  line  be  drawn  from  the  vertex  C  of  an  isosceles  triangle 
ABG,  meeting  the  base  AB  produced  at  D,  prgve  that 

CB'^-CB^  =  ADx  BD. 

54.  If  ^C  is  the  base  of  the  isosceles  triangle  ABC,  and  J.D  be 
drawn  perpendicular  to  BC,  prove  that 

3  AD^  +  2BD^  +  1J&  =  AB^  +  BC^  +  CA\ 

55.  The  middle  points  of  two  chords  are  distant  5  and  9  units, 
respectively,  from  the  middle  points  of  their  subtended  arcs.  If  the 
length  of  the  first  chord  is  20  units,  what  is  the  length  of  the  second  ? 

56.  The  sides  AB  and  BC  oi  &.  triangle  ABC  are  9  and  16, 
respectively,  and  the  length  of  the  median  drawn  from  A  is  11. 
Find  the  side  AC. 

57.  The  diameter  which  bisects  a  certain  chord  whose  length 
is  33|  units,  is  35  units  in  length.  Find  the  distance  from  either 
extremity  of  the  chord  to  the  extremities  of  the  diameter. 

58.  The  equal  angles  of  an  isosceles  triangle  are  each  30°,  and 
the  equal  sides  are  each  8  units  in  length.  What  is  the  length  of 
the  base  ? 

59.  The  diagonals  of  a  trapezoid,  whose  bases  are  AB  and  BC, 
intersect  at  E.    If  AE  =d,  EC=  3,  and  BD  =  IG,  find  BE  and  ED. 


\ 
156  PLANE   GEOMETRY.  —  BOOK  III. 

60.  Prove  the  theorem  of  §  284  by  drawing  A'B  and  AB\ 

61.  The  parallel  sides  of  a  circumscribed  trapezoid  are  6  and  18, 
respectively,  and  the  other  two  sides  are  equal  to  each  other.  Find 
the  radius  of  the  circle. 

62.  If  two  adjacent  sides  and  one  of  the  diagonals  of  a  parallel- 
ogram are  7,  9,  and  8,  respectively,  find  the  other  diagonal. 

63.  An  angle  of  a  triangle  is  acute,  right,;  or  obtuse  according  as 
the  square  of  the  opposite  side  is  less  than,  equal  to,  or  greater  than, 
the  sum  of  the  squares  of  the  other  two  sides. 

64.  Is  the  greatest  angle  of  a  triangle  whose  sides  are  3,  5,  and 
6,  acute,  right,  or  obtuse  ? 

65.  Is  the  greatest  angle  of  a  triangle  whose  sides  are  8,  9,  and 
12,  acute,  right,  or  obtuse  ? 

66.  Is  the  greatest  angle  of  a  triangle  whose  sides  are  12,  35,  and 
37,  acute,  right,  or  obtuse  ? 

67.  If  D  is  the  intersection  of  the  perpendiculars  from  the  vertices 
of  the  triangle  ABC  to  the  opposite  sides,  prove  that 

AB^  -  AC'^  =  B&  -  Clf. 

68.  If  a  parallel  to  the  hypotenuse  AB  of  the  right  triangle  ABC 
meets  AC  and  BC  at  D  and  jEJ,  prove  that 

Alf  +  B&  =  AB^  +  DE\ 

69.  The  diameters  of  two  circles  are  12  units  and  28  units, 
respectively,  and  the  distance  between  their  centres  is  29  units. 
Find  the  length  of  the  common  tangent  which  cuts  the  straight  line 
joining  the  centres. 

70.  State  and  prove  the  converse  of  Prop.  XXYI.,  II. 

71.  The  sum  of  the  squares  of  the  distances  of  any  point  in  the 
circumference  of  a  circle  from  the  vertices  of  an  inscribed  square, 
is  equal  to  twice  the  square  of  the  diameter  of  the  circle.     (§  196.) 

72.  The  sides  AB,  BC,  and  CA,  of  a  triangle  ABC,  are  169,  182, 
and  195  units,  respectively.  Find  the  segments  into  which  AB 
and  BC  are  divided  by  perpendiculars  drawn  from  C  and  A, 
respectively.     (  §  277. ) 

73.  In  a  right  triangle  ABC  is  inscribed  a  square  BEFG,  having 
its  vertices  D  and  G  in  the  hypotenuse  BC,  and  its  vertices  E  and  F 
in  the  sides  AB  and  AC.  Prove  that  BE  is  a  mean  proportional 
between  BB  and  CG. 


Note.    For  additional  exercises  on  Book  III.,  see  p.  224. 


SIMILAR  POLYGONS.  157 

PROBLEMS   IN  CONSTRUCTION. 

Proposition  XXXV.    Problem. 

289.    To  divide  a  given  straight  line  into  parts  propor- 
tional to  any  number  of  given  lines. 


Ji-^ 


Let  AB  be  the  given  straight  line. 

To  divide  AB  into  parts  proportional  to  the  given  lines 
Tn,  n,  and  p. 

Draw  the  indefinite  straight  line  AC. 

On  A  C  take  AD  =  m,  BE  =  n,  and  EF  =  p,  and  draw  BF. 

Draw  DG  and  EH  parallel  to  BF,  meeting  AB  at  G  and 
II,  respectively. 

Then  AB  is  divided  by  the  points  G  and  If  into  parts 
proportional  to  m,  n,  and  p. 

For  since  DG  is  parallel  to  the  side  EH  of  the  triangle 
AER, 

All         AG  GH  /(.   n4t;\ 

AE^Ab^DE'  (^^^^-^ 

That  is,                   AH^AG^GH  .^. 

'                    AE        m           n  ^    ^ 

And  since  EH  is  parallel  to  the  side  BF  of  the  triangle 

ABF, 

AE       EF       p    '  ^   ^ 

¥rom  (1;  and  (2), 

AG  ^  Gil ^  HB 
m  n  p 


158  PLANE   GEOMETRY.— BOOK  III. 


Proposition  XXXVI.     Problem. 

290.    To  divide  a  given  straight  line  into  any  number  of 
equal  parts. 


Let  AB  be  the  given  straight  line. 
To  divide  AB  into  four  equal  parts. 
Draw  the  indefinite  straight  line  AC. 
On  AC  take  any  convenient  length  AD,  and  lay  off  DU, 
EF,  and  FG  each  equal  to  AD. 
Dfaw  BG,  and  draw  DH,  EK,  and  FL  parallel  to  BG. 
Then,  AH  =  HK  =  KL  =  LB.  (§  243.) 

Proposition  XXXVII.     Problem. 

291.    To  construct  a  fourth  proportional  (§  230)  to  three 
given  straight  lines. 

' ?\ 


■" ^<  \ 

p  5^'   \ 

^         P       F.  G 

Let  m,  n,  and  p  be  the  given  straight  lines. 

To  construct  a  fourth  proportional  Jo  m,  n,  and  p. 

Draw  the  indefinite  straight  lines  AB  and  AC,  making 
any  convenient  angle  with  each  other. 

On  AB  take  AD  =  m  and  DE  =  n,  and  on  ^C  take 
AF  =p. 


SIMILAR  POLYGONS.  *  159 

Draw  DF,  and  draw  EG  parallel  to  DF. 
Then  FG  is  a  fourth  proportional  to  7/i,  n,  and  p. 
For  since  DF  is  parallel  to  the  side  l^G  of  the  triangle 
AEG, 

AD  :  DE  =  AF :  FG,  (§  245.) 

That  is,  m:n  =p  :  FG. 

292.  CoR.     If  AF  be  taken  equal  to  n,  the  above  pro- 
portion becomes 

m:  n  =  n:  FG. 

In  this   case,  FG  is  a  third  proportional  (§  229)  to  m 
and  ??/. 

Proposition  XXXVIII.     Problem. 

293.  To  construct  a  mean  proportional  (§  229)   between 
two  given  straight  lines. 

p,.,  y< 


A "^i  'B'"n  ""C E 

Let  m  and  n  be  the  given  straight  lines. 

To  construct  a  mean  proportional  between  m  and  n. 

On  the  indefinite  straight  line  AE,  take  AB  =  m  and 
BC  =  n. 

On  ^C  as  a  diameter  describe  the  semi-circumference 
ADC,  and  draw  BD  perpendicular  to  AC. 

Then  BD  is  a  mean  proportional  between  m  and  n. 

*For,  AB:BD  =  BD:  BC.  (§  272,  I.) 

That  is,  m:BD  =  BD:n. 

294.  ScH.  By  aid  of  §  293,  a  line  may  be  constructed 
equal  to  Va,  where  a  is  any  number  whatever. 

Thus,  to  construct  a  line  equal  to  V3,  we  take  AB  equal 
to  3  units,  and  BC  equal  to  1  unit. 

Then,  BD  =  VAB  X  BC  (§  231)  =  VS^O  =  VS. 


160 


PLANE   GEOMETKY.  —  BOOK  III. 


295.  Def.  a  straight  line  is  said  to  be  divided  by  a 
given  point  in  extreme  and  mean  ratio  when  one  of  the 
segments  (§  251)  is  a  mean  proportional  between  the  whole 
line  and  the  other  segment. 


B 


Thus,  the  line  AB  is  divided  internally  in  extreme  and 
mean  ratio  at  the  point  C  if 

AB:AC  =  AC:BC) 
and  it  is  divided  externally  in  extreme  and  mean  ratio  at 
the  point  Z>  if 

AB:AD  =  AD'.  BD. 


Proposition  XXXIX.     Problem. 

296.    To  divide  a  given  straight  line  in  extreme  and  mean 
ratio  (§  295). 


X 


/  . 


F^ 


D  A.  CB 

Let  AB  be  the  given  straight  line. 

To  divide  AB  in  extreme  and  mean  ratio. 

Draw  BE  perpendicular  to  AB,  and  equal  to  ^  AB. 

With  ^  as  a  centre,  and  EB  as  a  radius,  describe  the 
circumference  BFG. 

Draw  the  straight  line  AU  cutting  the  circumference  at 
F  and  G. 

On  AB  take  AC  =  AF;  on  BA  produced  take  AD  =  AG. 

Then  AB  is  divided  at  C  internally,  and  at  D  externally, 
in  extreme  and  mean  ratio. 


SIMILAR  POLYGONS. 

For* since  AG  is  a,  secant,  and  AB  a  tangent, 

AG:AB  =  AB:AF.  (§  283.) 

That  is,  AG.AB  =  AB:AC.  (1) 

Then,         AG-AB:AB  =  AB-AC:AC.         (§237.) 

Whence,    AB  :AG  -  AB  =  AC :  BC.  (§  235.) 

But  by  hypothesis, 

AB  =  2  BE  =  FG.  (2) 

Whence,  AG  -  AB  =  AG  -  FG 

=  AF  =  Aa 

Substituting,  we  have 

AB:AC  =  AC:BC.  (3) 

Therefore  AB  is  divided  at  C  internally  in  extreme  and 
mean  ratio. 

Again,  from  (1), 

AG-\-AB:AG  =  AB-\-  AC.AB.        (§  236.) 

But,  AG  +  AB  =  AD-{-AB  =  BD. 

And  by  (2),       AB  +  AC  =  FG -\- AF  =  AG. 

Therefore,  BD :  AG  =  AG  :  AB. 

Whence,  AB  :  AG  ^  AG  :  BD.  (§  235.) 

That  is,  AB:AD  =  AD:  BD. 

Therefore  AB  is  divided  at  D  externally  in  extreme  and 
mean  ratio. 

297.   CoR.     If  AB  be  denoted  by  m,  and  AC  by  ic,  the 
proportion  (3)  of  the  preceding  article  becomes 
mix  =  x:m  —  X. 

Placing  the  product  of  the  means  equal  to  the  product  of 
the  extremes,  we  have 

x^  =  wi2  _  ^^^  ^§  231.) 

or,  £c^  +  mx  =  m\ 

Solving  this  quadratic  equation,  we  obtain 


Ex.  74.   Construct  a  line  equal  to  V2  ;  to  VE  ;  to  Vq. 


162  PLANE   GEOMETRY.  — BOOK  III. 

Proposition  XL.     Pkoblem. 

298.  Upon  a  given  side,  homologous  to  a  given  side  of  a 
given  polygon,  to  construct  a  polygon  similar  to  the  given 
polygon. 


Let  ABODE  be  the  given  polygon,  and  A'B'  the  given 
side. 

To  construct  upon  the  side  A'B',  homologous  to  AB,  a 
polygon  similar  to  ABODE. 

Divide  the  polygon  ABODE  into  triangles  by  drawing 
the  diagonals  EB  and  EO. 

At  A'  construct  Z  B'A'E'  =  Z.  A,  and  draw  B'E'  making 
ZA'B'E'  =AABE. 

Then  the  triangle  A'B'E'  wiU  be  similar  to  ABE.  (§  256.) 

In  like  manner,  construct  the  triangle  B'O'E'  similar  to 
BOE,  and  the  triangle  O'D'E'  similar  to  ODE. 

Then  AB' O'D'E'  will  be  similar  to  ABODE. 

For  two  polygons  are  similar  when  they  are  composed  of 
the  same  number  of  triangles,  similar  each  to  each,  and 
similarly  placed  (§  266). 

EXERCISES. 

75.  To  inscribe  in  a  given  circle  a,  triangle  similar  to  a  given 
triangle. 

76.  To  circumscribe  about  a  given  circle  a  triangle  similar  to  a 
given  triangle. 

Note.    For  additional  exercises  on  Book  III.,  see  p.  226. 


BOOK   IV. 


AREAS    OF    POLYGONS. 


Proposition  I.     Theorem. 

299.    Two  rectangles  having  equal  altitudes  are  to  each 
other  as  their  bases. 

Note.     The  word  ^^rectanyle^''  in  the  above  statement,  signifies 
the  amount  of  surface  of  the  rectangle. 

Case  I.    Whe?i  the  bases  are  eommensurable. 


E 


C 


A     Q 


D 


Let  ABCD  and  ABEF  be  two  rectangles,  having  equal 
altitudes,  and  commensurable  bases  AD  and  AF. 

,p  ABCD      AD 

lo  prove  = . 

^  ABEF      AF 

Let  ^(t  be  a  common  measure  of  AD  and  AF,  and  let  it 
be  contained  7  times  in  AD,  and  4  times  in  AF. 
AD  ^7 

AF~   4:' 

Drawing  perpendiculars  to  AD  througli  the  several  points 
of  division,  the  rectangle  ABCD  will  be  divided  into  7  equal 
parts,  of  which  ABFF  will  contain  4.  (§  113.) 

ABCD  ^  7 
ABFF      4 ' 
ABCD  ^  AD 
ABFF      AF ' 
163 


Then, 


Then, 

From  (1)  and  (2), 


(1) 


(2) 


164  PLANE   GEOMETRY. —BOOK  IV. 

Case  II.    When  the  bases  are  incommensurable. 
B  a   E  o 


Let  ABCD  and  ABEF  be  two  rectangles,  having  equal 
altitudes,  and  incommensurable  bases  AD  and  AF. 
ABGD^AD 
AF' 


To  prove 


ABEF 

Let  AD  be  divided  into  any  number  of  equal  parts,  and 
let  one  of  these  parts  be  applied  to  AF  as  a  measure. 

Since  AD  and  AF  are  incommensurable,  a  certain  num- 
ber  of  the   parts   will   extend   from   A   to   H,   leaving   a 
remainder,  HF,  less  than  one  of  the  parts. 
Draw  GH  perpendicular  to  AD. 
ABCD  ^  AD 
AH 


Then, 


(§  299,  Case  I.) 


ABGH 

Now  let  the  number  of  subdivisions  of  AD  be  indefinitely 
increased. 

Then  the  length  of  each  part  will  be  indefinitely  di- 
minished, and  the  remainder  HF  will  approach  the  limit  0. 


Then, 


ABCD     ...  .   , ,     T    .^ABCD 

will  approach  the  limit 


ABGH 


and 


AD 

-——  will  approach  the  limit 
AH  ^^  AF 


ABEF' 
AD 


By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 

Whence,  A^CD^AD 

ABEF      AF 

300.  CoR.  Since  either  side  of  a  rectangle  may  be  taken 
as  the  base,  it  follows  that 

Two  7'ectangles  having  equal  bases  are  to  each  other  as 
their  altitudes. 


AREAS   OF  POLYGONS. 


165 


Proposition  II.     Theorem. 

301.   Any  two  rectangles  are  to  each  other  as  the  products 
of  their  bases  hy  their  altitudes. 


«! 


b' 


Let  A  and  B  be  any  two  rectangles,  having  the  altitudes 
a  and  a',  and  the  bases  b  and  b',  respectively. 

A        aXb 
B 


To  prove 


a'X  V 

Let  C  be  a  rectangle  having  the  altitude  a  and  the  base  b'. 

Then  the  rectangles  A  and  C,  having  equal  altitudes,  are 

to  each  other  as  their  bases.  (§  299.) 

That  is,  i  =  |.  (1) 

And  the  rectangles  C  and  B^  having  equal  bases,  are  to 
each  other  as  their  altitudes.  (§  300.) 

That  is,  ^  =  ±-. 

'  B       a! 

Multiplying  (1)  and  (2), 

A^  _  a  Xb 

B  ~  a'X  b'' 


(2) 


302.  Sen.  It  must  be  observed  that  the  product  of  tiro 
lines  signifies  the  product  of  their  numerical  measures  when 
referred  to  a  common  unit. 


Ex.  1.  A  rectangular  field  is  182  feet  long,  and  102  feet  wide. 
Another  is  119  feet  long,  and  117  feet  wide.  Wliat  is  the  ratio  of 
their  surfaces  ?  . 


166 


PLANE  GEOMETEY.  —  BOOK  IV. 


DEFINITIONS. 

303.  The  area  of  a  surface  is  its  ratio  to  another  surface, 
called  the  unit  of  surface,  adopted  arbitrarily  as  the  unit  of 
measure  (§  179). 

Thus,  if  A  represents  a  certain  surface,  and  B  the  unit  of 

surface,  the  area  of  ^  is  — - . 

The  usual  unit  of  surface  is  the  square  tvhose  side  is  some 
linear  unit ;  for  example,  a  square  inch,  or  a  square  foot. 

304.  Two  surfaces  are  said  to  be  equivalent  when  their 
areas  are  equal. 

The  symbol  =o=  is  used  for  the  words  "  is  equivalent  to." 

Proposition  III.     Theorem. 

305.  The  area  of  a  rectangle  is  equal  to  the  product  of 
its  base  and  altitude. 


B 


Let  a  and  h  be  the  altitude  and  base  of  the  rectangle  A ; 
and  let  B  be  the  unit  of  surface ;  i.e.,  a  square  whose  side 
is  the  linear  unit. 

To  prove  area  oi  A  =  a  X  h. 

Any  two  rectangles  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes  (§  301). 
A_  _  a  X  b 
B  ~1X1 


Whence, 


=  aX  b. 


But  since  B  is  the  unit  of  surface,  —  is  the  area  of  A. 

(§  303.) 
Therefore,  area  oi  A  =  a  x  b. 


AREAS  OF  POLYGONS.  167 

306.    Cor.    The  area  of  a  square  is  equal  to  the  square  of 

its  side. 

2ICn.  Sen.  I.  The  statement  of  Prop.  III.  is  an  abbrevia- 
tion of  the  following : 

If  the  unit  of  surface  is  the  square  whose  side  is  the 
linear  unit,  the  number  which  expresses  the  area  of  a  rect- 
angle is  equal  to  the  product  of  the  numbers  which  express 
the  lengths  of  its  sides. 

An  interpretation  of  the  above  form  is  always  understood 
in  every  proposition  relating  to  areas. 

308.  ScH.  II.  If  the  sides  of  the  rectangle  are  multiples 
of  the  linear  unit,  tlie  truth  of  Prop.  III. 

may  be  seen  by  dividing  the  figure  into 
squares,  each  equal  to  the  unit  of  sur- 
face. 

Thus,  if  the  altitude  of  the  rectangle  A 
is  5  units,  and  its  base  6  units,  the  figure 
can  evidently  be  divided  into  30  squares. 

In  this  case,  30,  the  number  which  expresses  the  area  of 
the  rectangle,  is  the  product  of  6  and  5,  the  numbers  which 
express  the  lengths  of  the  sides. 

309.  Def.  The  dimensions  of  a  rectangle  are  its  base  and 
altitude. 

EXERCISES. 

2.  If  the  area  of  a  rectangle  is  7956  sq.  in.,  and  its  base  S^-  yd., 
find  its  perimeter  in  feet. 

3.  If  the  base  and  altitude  of  a  rectangle  are  14  ft.  7  in.,  and  5  ft. 
3  in.,  respectively,  what  is  the  side  of  an  equivalent  square  ? 

4.  Find  the  dimensions  of  a  rectangle  whose  area  is  168,  and. 
perimeter  52. 

5.  The  area  of  a  rectangle  is  143  sq.  ft.  75  sq.  in.,  and  its  base  is 
3  times  its  altitude.     Find  each  of  its  dimensions. 

6.  The  area  of  a  square  is  693  sq.  yd.  4  sq.  ft. ;  find  its  side. 


— 1. — i — 4 — I — 4— 

!       '       !       I        I 

. — I — 4 — ;...4. — i— 

...1.4-4-l-i- 


168 


PLANE  GEOMETRY.  —  BOOK  IV. 


Proposition  IV.     Theorem. 

310.    The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  ba^e  and  altitude. 

E       B F        0 


Let  ABCD  be  a  parallelogram,  hating  its  altitude  DF 
equal  to  a,  and  its  base  AD  equal  to  b. 
To  prove  area  ABCD  =  a  X  b. 

Draw  AU  perpendicular  to  AD,  meeting  CB  produced  at 
FJ,  forming  the  rectangle  AEFD. 

Then  in  the  right  triangles  ABE  and  DCF,    ' 

AB  =  DC,  and  AF  =  DF.  (§  104.) 

Whence,  A  ABF  =  A  DCF.  (§  88.) 

Now,  if  from  the  entire  figure  AECD  we  take  the  tri- 
angle ABE,  there  remains  the  parallelogram  ABCD)  and 
if  we  take  the  triangle  DCF,  there  remains  the  rectangle 
AEFD. 

Therefore,        area  ABCD  =  area  AEFD. 

But,  area  AEFD  ^  axb.  (§  305.) 

Whence,  area  ABCD  =  aXb. 

311.  Cor.  I.  Two  parallelograms  having  equal  bases  and 
equal  altitudes  are  equivalent  (§  304). 

312.  Cor.  II.  1.  Two  parallelograms  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 

2.  Two  parallelograms  having  equal  bases  are  to  each 
other  as  their  altitudes. 

3.  Any  two  parallelograms  are  to  each  other  as  the  prod- 
ucts of  their  bases  by  their  altitudes. 


AREAS   OF  POLYGONS.  169 


Proposition  V.    Theorem. 

313.    The  area  of  a  triangle  is  equal  to  one-half  the  prod- 
uct of  its  base  and  altitude. 


Let  ABC  be  a  triangle,  having  its  altitude  AE  equal  to 
a,  and  its  base  BC  equal  to  h. 

To  prove  area  ABC  =  \a  xh. 

Draw  AD  and  CD  parallel  to  BC  and  AB,  respectively, 
forming  the  parallelogram  ABCD. 

Now,  AABC=  AACD.  (§  106.) 

Whence,  area  ABC  =  ^  area  ABCD. 

But,  area  ABCD  =  aXb.  (§  310.) 

Whence,  area  ABC  =^a  xb. 

314.  CoR.  I.    Two  triangles  having  equal  bases  and  equal 
altitudes  are  equivalent. 

315.  CoR.  II.    1.    Two  triangles  having  equal  altitudes 
are  to  each  other  as  their  bases. 

2.  Two  triangles  having  equal  bases  are  to  each  other  as 
their  altitudes. 

3.  Ang  two  triangles  are  to  each  other  as  the  pi'oducts  of 
their  bases  by  their  altitudes. 

316.  Cor.  III.    A  triangle  is  equivalent  to  one-half  of  a 
parallelogram  having  the  same  base  and  altitude. 


Ex.  7.   The  hypotenuse  of  a  right  triangle  is  5  ft.  5  in.,  and  one  of 
its  legs  is  2  ft.  9  in.     Find  its  area. 


170  PLANE  GEOMETRY.  —  BOOK  IV. 


Proposition  VI.     Theorem. 

317.    The  area  of  a  trapezoid  is  equal  to  one-half  the  sum 
of  its  bases  multiplied  by  its  altitude. 


D 

h 

C 

/f^ 

\ 

fl/.  .  y 

__X 

/  f 

\^ 

•4 

E 


Let  ABCD  be  a  trapezoid,  having  its  altitude  DE  equal 
to  a,  and  its  bases  AB  and  DC  equal  to  h  and  b'^  respectively. 
To  prove  area  ABCD  =  a  X  ^  (b -\-  b'). 

Draw  the  diagonal  BD. 

Then  the  trapezoid  is  divided  into  two  triangles,  ABD 
and  BCD,  having  the  common  altitude  a. 

Whence,  area  ABD  =  ^a  X  b, 

and  area  BCD  =  ^aXb'.  (§  313.) 

Adding,  we  have  ^ 

area  ABCD  =  ^aXb-{-^aXb'          1 
=  aX^(b-\-b'). 

318.  CoR.  Since  the  line  joining  the  middle  points  of 
the  non-parallel  sides  of  a  trapezoid  is  equal  to  one-half  the 
sum  of  the  bases  (§  132),  it  follows  that 

The  area  of  a  trapezoid  is  equal  to  the  product  of  its 
altitude  by  the  line  joining  the  middle  points  of  its  non- 
parallel  sides. 

319.  Sch:  The  area  of  any  polygon  may  be  obtained  by 
finding  the  sum  of  the  areas  of  the  triangles  into  which  the 
polygon  may  be  "divided  by  drawing  diagonals  from  any  one 
of  its  vertices. 


AREAS   OF  POLYGONS. 


171 


But  in  practice  it  is  better  to  draw  the  longest  diagonal, 
and  draw  perpendiculars   to  it   from 
the  remaining  vertices  of  the  polygon. 

The  polygon  will  then  be  divided 
into  right  triangles  and  trapezoids  ; 
and  by  measuring  the  lengths  of  the 
jjerpendiculars,  and  of  the  portions 
of  the  diagonal  which  they  intercept, 
the  areas  of  the  figures  inay  be  found  by  §§  313  and  317. 


Proposition  VII.    Theorem. 


320.    Two   siinilar    triangles  are    to   each   other  as   the 
squares  of  their  homologous  sides. 


Let  AB  and  A'B'  be  homologous  sides  of  the   similar 
triangles  ABC  and  A'B'C\ 


To  prove 


ABC 
A'B'C' 


AB' 


A'B' 

Draw  the  altitudes  CD  and  CD'. 


Then, 


But, 


ABC 


AB  X  CD 


A'B'C       A'B'  X  CD' 
A'B'        CD' 


CD 


AB 


CD'        A'B' 
Substituting  this  value  in  ( 1 ),  we  have 

ABC    '     AB    ^,  AB         AB^ 


A'B'C       A'B'       A'B'      j^' 


(§  315,  3.> 

(1) 
(§  263.) 


17^  PLANE  GEOMETRY.  —  BOOK  IV. 

321.  ScH.  Two  similar  triangles  are  to  each  other  as 
the  squares  of  any  two  homologous  lines. 

Proposition  VIII.     Theorem. 

322.  Two  triangles  having  an  angle  of  one  equal  to  an 
angle  of  the  other,  are  to  each  other  as  the  products  of  the 
sides  including  the  equal  angles. 


Let  the  triangles  ABC  and  AB'C  have  the  common 
angle  A. 

rr.  ABC         ABXAC 

To  prove  ___  =  ______. 

Draw  B'C 

Then  the  triangles  ABC  and  AB'C,  having  the  common 
vertex  C,  and  their  bases  AB  and  AB'  in  the  same  straight 
line,  have  the  same  altitude. 

Whence,  ^  =  ^-  (§315,1.) 

T    ri  ^^'0        AC 

In  like  manner. 


AB'C      AC 
Multiplying  these  equations,  we  have 

ABC  ^  ABXAG 
AB'C      AB'XAC' 

EXERCISES. 

8.  If  the  altitude  of  a  trapezoid  is  1  ft.  4  in.,  and  its  bases  are 
1  ft.  1  in.  and  2  ft.  5  in.,  respectively,  what  is  its  area  ? 

9.  If,  in  the  figure  of  Prop.  VII.,  AB  =  9,  A'B'  =  7,  and  the  area 
of  A'B'Cf  is  147,  what  is  the  area  of  ABC  ? 


AREAS  OF  POLYGONS.  173 


Proposition  IX.     Theorem. 

323.    Two  similar  polygons  are  to  each  other  as  the  squares 

of  their  homologous  sides. 


Let  AB  and  A'B'  be  homologous  sides  of  the  similar 
polygons  AC  and  A'C,  whose  areas  are  K  and  K'j  re- 
spectively. 


K         A& 


Draw  the  diagonals  EB,  EC,  E'B',  and  E'C\ 

Then  the  triangle  ABE  is  similar  to  A'B'E'.        (§  267.) 

ABE  AB" 

Whence,  ^7^7^  =  -=;-,.  (§  320.) 

In  like  manner, 


and 


A'B'E'       A'B 

/2 

iier, 

BCE 

BC'' 

AB' 

B'C'E' 

B'C' 

A'B''  ' 

CDE 

CD' 

AB' 

C'D'E' 

CD'' 

A'B'' 

ABE 

BCE 

:  ■ ■    r= 

CDE 

Whence  ^^^     —    ^^-^     —     "^^^  rXy-  1  ^ 

'       TWW  ~  WWW'  ~  WWW  '  ^^'^'  ^'^ 

Then  ABE  ^  BCE -\- CDE         ^    ABE 

'  A'B'E''-}-B'C'E'  +  C'D'E'      A'B'E'' 

(§  239.) 

Therefore,  K_^ABE_^AB^ 

'  K'      A'B'E'      J4^2 

324.    ScH.   Two  similar  polygons  are  to  each  other  as  the 
squares  of  any  two  homologous  lines. 


174 


PLANE  GEOMETRY.— BOOK  IV. 


Proposition  X.    Problem. 

325.    To  find  the  area  of  a  triangle  when  its  three  sides 
are  given. 


Let  a,  b,  and  c  represent  the  sides  BC,  CA,  ^d  AB,  and 
^the  area,  of  the  triangle  ABC. 
To  find  K  in  terms  of  a,  h,  and  c. 

Let  C  be  an  acute  angle,  and  draw  AD  perpendicular 
to  BC. 

Then,  c'' =  a'' -^  h'' —  2  a  X  CD.       (§  277.) 

Transposing,     2aXCD  =  a"" -]-h''  — g\ 

Whence, 

Then, 


2a 


ID'  =  AC'-  CD' 


(§  274.) 


(AC -\- CD)  (AC -  CD) 


b  + 


2a 


^2  _|.  ^2  _  ^2\ 

2a         ) 


_(2ab  +  a^-^b^  —  c^)  (2  ab  -  a' -  b^  +  c^) 

4.  a'' 
_  [(a  -I-  by  -  c^-]  [c^  -  (a-  b)^-] 

4.a^ 
^(a  ^h  -\-  c)  (a  -\-b  —  r)  (c  -\-  a  —  b)  (c  —  a  -\-h) 
4.  a' 

Now  let  a  -\-  b  -\-  G  =  2  s. 
Then, 
a-^b  —  c  =  a-\-b-\-c  —  2c  =  2s~2G  =  2(s  —  c). 


(1) 


AREAS  OF  POLYGONS.  175 

In  like  manner, 

c  -{-  a  —  b  =  2  (s  —  b),  and  c  —  a  -\-  b  =  2  (s  —  a). 
Substituting  in  (1),  we  have 


Whence, 


J  J  ^  2  V '''•  (s  -a)(s-  b)  {s  -  c) 


Then,  K=ia  xAD  (§  313.) 

=  V^  {s  —  a)  (s  —  b)  (s  —  c). 

EXERCISES. 
10.  The  sides  of  a  triangle  are  13,  14,  and  15 ;  find  its  area. 

SOLUTION. 

Let  a  =  13,  b  =  14,  and  c  =  15. 
Then,  s  =  i  (13  +  14  +  15)  =  21. 
Whence,  s  —  a  =  8,  s  —  6  =  7,  and  s  —  c  =6. 
Therefore,  the  area  of  the  triangle  is 

V21  X  8  X  7  X  6  =  V3x7x23x7x2x3 


=  V2*  x  32  X  72 

=  22  x  3  X  7  =  84,  Ans. 

11.  If  the  sides  of  a  triangle  ABC  are  AB  =  25,  BC  =  17,  and 
CA  =  28,  find  its  area,  and  the  length  of  the  perpendicular  from 
each  vertex  to  the  opposite  side. 

12.  Find  the  length  of  the  diagonal  of  a  rectangle  whose  area  is 
2040,  and  altitude  48. 

13.  Find  the  lower  base  of  a  trapezoid  whose  area  is  9408,  upper 
base  79,  and  altitude  96. 

14.  The  area  of  a  rhombus  is  equal  to  one-half  the  product  of  its 
diagonals. 

15.  The  diagonals  of  a  parallelogram  divide  it  into  four  equivalent 
triangles. 

16.  Lines  drawn  to  the  vertices  of  a  parallelogram  from  any  point 
in  one  of  its  diagonals  divide  the  figure  into  two  pairs  of  equivalent 
triangles. 

17.  If  EF  is  any  straight  line  drawn  through  the  centre  of  the 
parallelogram  ABCD,  meeting  the  sides  AD  and  BC  at  E  and  F, 
then  the  triangles  BFF  and  CED  are  equivalent. 


176 


PLANE  GEOMETRY.  — BOOK  lY. 


326.  Since  the  area  of  a  square  is  equal  to  the  square  of 
its  side  (§  306),  we  may  state  Prop.  XXVII.,  Book  III.,  as 
follows : 

In  any  right  triangle,  the  square  described  upon  the 
hypotenuse  is  equivalent  to  the  sum  of  the  squares  described 
upon  the  legs. 

The  theorem  in  the  above  form  is  proved  as  follows  : 


Let  AB  be  the  hypotenuse  of  the  right  triangle  ABC. 

To  prove  that  the  square  ABEF*  described  upon  AB,  is 
equivalent  to  the  sum  of  the  squares  AC  GIT  and  BCKL, 
described  upon  AC  and  BC,  respectively. 

Draw  CD  perpendicular  to  AB. 

Produce  CD  to  meet  EF  at  M,  and  draw  BH  and  CF. 

Then  in  the  triangles  ABH  and  A  CF, 

AB  =  ^i^,^and  AH  =  AC. 

Also,  Z  BAH  =  Z  CAF, 

since  each  is  equal  to  a  right  angle  +  Z  BAC. 

Hence,  A  ABH=  A  ACF.  (§  63.) 

Now  the  triangle  ABH  has  the  same  base  and  altitude  as 
the  square  ACGH. 

Whence,  area  ABH  =  ^  area  ACGH.  (§  316.) 

Again,  the  triangle  ^Ci^has  the  same  base  and  altitude 
as  the  rectangle  ADMF. 


AREAS  OF  POLYGONS.  177 

Whence,  area  ACF  =  \ area  ADMF. 

But,  area  ABH  =  area  A  CF. 

•Whence,      -    Sivesi  AC GH  ==  area  ADMF.  (1) 

In  like  manner,  by  drawing  AL  and  CFj  we  may  prove 

area  BCKL  =  area  BDME.  (2) 

Adding  (1)  and  (2),  we  have 

area  ACGH -\-  area  BCKL  =  3LTea.ABFF, 

327.  ScH.  The  above  theorem  is  supposed  to  have  been 
first  given  by  Pythagoras,  and  is  called  after  him  the  Pytha- 
gorean Theorem. 

Several  other  propositions  of  Book  III.  may  be  put  in  the 
form  of  statements  in  regard  to  areas;  as,  for  example, 
Props.  XXVIII.  and  XXIX. 

EXERCISES. 

18.  The  side  of  an  equilateral  triangle  is  5  ;  find  its  area. 

19.  The  altitude  of  an  equilateral  triangle  is  3;  find  its  area. 

20.  The  area  of  a  certain  triangle  is  2^  times  the  area  of  a  similar 
triangle.  If  the  altitude  of  the  first  triangle  is  4  ft.  3  in.,  what  is  the 
homologous  altitude  of  the  second  ? 

21.  Two  triangles  are  equivalent  when  they  have  two  sides  of  one 
equal  respectively  to  two  sides  of  the  other,  and  the  included  angles 
supplementary. 

22.  One  diagonal  of  a  rhombus  is  five-thirds  of  the  other,  and  the 
difference  of  the  diagonals  is  8;  find  its  area. 

23.  If  B  and  E  are  the  middle  points  of  the  sides  BC  and  ^C  of 
the  triangle  ABC^  prove  that  the  triangles  ABD  and  ABE  are 
equivalent. 

24.  If  E  is  the  middle  point  of  CD,  one  of  the  non-parallel  sides 
of  the  trapezoid  ABCD,  and  a  parallel  to  AB  drawn  through  E 
meets  BC  Sit  F  and  AD  at  G,  prove  that  the  parallelogram  ABFG 
is  equivalent  to  the  trapezoid. 

25.  The  sides  AB,  BC,  CD,  and  DA  of  the  quadrilateral  ABCD 
are  10,  17,  13,  and  20,  respectively,  and  the  diagonal  J.  C  is  21.  Find 
the  area  of  the  quadrilateral. 

26.  Find  the  area  of  the  square  inscribed  in  a  circle  whose  radius 
is  3. 


178  PLANE  GEOMETKY.  —  BOOK  IV. 

27.  The  area  of  an  isosceles  right  triangle  is  81  sq.  in. ;  find  its 
hypotenuse  in  feet. 

28.  The  area  of  an  equilateral  triangle  is  dVS;  find  its  side.     • 

29.  The  area  of  an  equilateral  triangle  is  16  V3 ;  find  its  altitude. 

30.  The  base  of  an  isosceles  triangle  is  56,  and  each  of  the  equal 
sides  is  53 ;  find  its  area. 

31.  The  area  of  a  triangle  is  equal  to  one-half  the  product  of  its 
perimeter  by  the  radius  of  the  inscribed  circle. 

32.  The  area  of  an  isosceles  right  triangle  is  equal  to  one-fouith 
the  area  of  the  square  described  upon  the  base. 

33.  If  the  angle  A  of  the  triangle  ABC  is  30°,  prove  that 

area.  ABC  =  iAB  X  AC. 

34.  A  circle  whose  diameter  is  12  is  inscribed  in  a  quadrilateral 
whose  perimeter  is  40.     Find  the  area  of  the  quadrilateral. 

35.  Two  similar  triangles  have  homologous  sides  equal  to  8  and 
15,  respectively.  Find  the  homologous  side  of  a  similar  triangle 
equivalent  to  their  sum. 

36.  The  non-parallel  sides  of  a  trapezoid  are  each  25  units  in 
length,  and  the  parallel  sides  are  19  and  33  units,  respectively. 
Find  the  area  of  the  trapezoid. 

37.  If  E  is  any  point  within  the  parallelogram  ABCD,  the 
triangles  ABE  and  CBE  are  together  equivalent  to  one-half  the 
parallelogram. 

38.  If  the  area  of  a  polygon,  one  of  whose  sides  is  15  in.,  is  375 
sq.  in.,  what  is  the  area  of  a  similar  polygon  whose  homologous  side 
is  18  in.  ? 

39.  If  the  area  of  a  polygon,  one  of  whose  sides  is  36  ft.,  is  648 
sq.  ft.,  what  is  the  homologous  side  of  a  similar  polygon  whose  area 
is  392  sq.  ft.? 

40.  If  one  diagonal  of  a  quadrilateral  bisects  the  other,  it  divides 
the  quadrilateral  into  two  equivalent  triangles. 

41.  Two  equivalent  triangles  have  a  common  base,  and  lie  on 
opposite  sides  of  it.  Prove  that  the  base,  produced  if  necessary, 
bisects  the  line  joining  their  vertices. 

42.  If  the  sides  of  a  triangle  are  15,  41,  and  52,  find  the  radius  of 
the  inscribed  circle.     (Ex.31.) 

43.  The  area  of  a  rhombus  is  240,  and  its  side  is  17;  find  its 
diagonals.  •» 


AREAS   OF  POLYGONS.  179 

44.  If  the  sides  of  a  triangle  are  25,  29,  and  36,  find  the  diameter 
of  the  circumscribed  circle.      (§  287.) 

45.  The  smn  of  the  perpendiculars  from  any  point  within  an 
equilateral  triangle  to  the  three  sides  is  equal  to  the  altitude  of  the 
triangle. 

46.  The  longest  sides  of  two  similar  polygons  are  18  and  3,  re- 
spectively. How  many  polygons,  each  equal  to  the  second,  will  form 
a  polygon  equivalent  to  the  first  ? 

47.  The  sides  AB  and  AC  of  a  triangle  ABC  are  15  and  22, 
respectively.  From  a  point  D  in  AB,  &  parallel  to  BC  is  drawn 
meeting  AC  &t  E,  and  dividing  the  triangle  into  two  equivalent 
parts.     Find  AB  and  AE. 

48.  The  segments  of  the  hypotenuse  of  a  right  triangle  made  by 
a  perpendicular  drawn  from  the  vertex  of  the  right  angle,  are  5§  and 
9f ;  find  the  area  of  the  triangle. 

49.  Any  straight  line  drawn  through  the  centre  of  a  parallelo- 
gram, terminating  in  a  pair  of  opposite  sides,  divides  the  paral- 
lelogram into  two  equivalent  quadrilaterals. 

50.  If  E  is  the  middle  point  of  CB,  one  of  the  non-parallel  sides 
of  the  trapezoid  ABCB,  prove  that  the  triangle  ABE  is  equivalent 
to  one-half  the  trapezoid. 

51.  The  sides  of  a  triangle  ABC  are  AB  =  13,  BC=  14,  and 
CA  =  15.  If  J  D  is  the  bisector  of  the  angle  A,  find  the  areas  of 
the  triangles  ABB  and  ACB. 

52.  The  longest  diagonal  AB  of  a  pentagon  ABCBE  is  44,  and 
the  perpendiculars  to  it  from  B,  C,  and  E  are  24,  16,  and  15,  re- 
spectively. If  AB  =  25,  CB  =  20,  and  AE  =  17,  what  is  the  area 
of  the  pentagon  ? 

53.  The  sides  of  a  triangle  are  proportional  to  the  numbers  7,  24, 
and  25.  The  perpendicular  to  the  third  side  from  the  vertex  of  the 
opposite  angle  is  13^-^.     Find  the  area  of  the  triangle. 

54.  If  E  and  F  are  the  middle  points  of  the  sides  AB  and  ^C  of 
a  triangle,  and  B  is  any  point  in  J5C,  prove  that  the  quadrilateral 
AEBF  is  equivalent  to  one-half  the  triangle  ABC. 

55.  The  parallelogram  formed  by  joining  the  middle  points  of 
the  adjacent  sides  of  a  quadrilateral  is  equivalent  to  one-half  the 
quadrilateral. 

Note.     For  additional  exercises  on  Book  TV.,  see  p.  226. 


180  PLANE  GEOMETRY.  —  BOOK  IV. 

PROBLEMS  IN   CONSTRUCTION. 
Proposition  XI.     Problem. 
328.    To  construct  a  triangle  equivalent  to  a  given  polygon. 


Let  ABODE  be  the  given  polygon. 

To  construct  a  triangle  equivalent  to  ABODE. 

Let  Af  B,  and  0  be  any  three  consecutive  vertices,  and 
draw  the  diagonal  AO. 

Draw  BF  parallel  to  AO,  meeting  DO  produced  at  F, 
and  draw  AF. 

Then  AFDE  is  a  polygon  equivalent  to  ABODE,  having 
a  number  of  sides  less  by  one. 

For  the  triangles  ABO  and  AFO  have  the  same  base  AO. 

And  since  their  vertices  B  and  F  lie  in  the  same  straight 
line  parallel  to  AO,  they  have  the  same  altitude.         (§  78.) 

Therefore,  area  ABO  =  area  AFO.  (§  314.) 

Adding  area  AODE  to  both  members,  we  have 
area  ABODE  =  area  AFDE. 

Again,  draw  the  diagonal  AD. 

Draw  EG  parallel  to  AD,  meeting  OD  produced  at  G, 
and  draw  AG. 

Then,  area  AED  =  area  AGD.  (§  314.) 

Adding  area  AFD  to  both  members,  we  have 
area  AFDE  =  area  AFG. 

Whence,       area  ABODE  =  area  AFG.  (Ax.  1.) 


AREAS  OF  POLYGONS. 


181 


Proposition  XII.     Problem. 

329.    To  construct  a  square  equivalent  to  the  sum  of  two 
given  squares. 


• 

1  \ 

i- — --& 

p 

M 

N 

Let  M  and  iV  be  tlie  given  squares. 

To  construct  a  square  equivalent  to  the  sum  of  M  and  N. 

Draw  AB  equal  to  a  side  of  Jtf ;  at  ^  draw  AC  perpen- 
dicular to  AB,  and  equal  to  a  side  of  N,  and  draw  BC. 

Then  the  square  P,  described  with  its  side  equal  to  BC, 
will  be  equivalent  to  the  sum  of  3f  and  iV. 

Por  in  th«  right  triangle  ABC, 

(§  273.) 


BC'==AB'-j-AC\ 


That  is, 


area  P  =  area  M  -f-  area  N. 


(§  306.) 


P/ 


Ok 


330.   GoR.   By  an  extension  of  the  above  method,  a  square 
may  be  constructed  equivalent  to  the  sum  of 
any  number  of  given  squares. 

Let  it  be  required,  for  example,  to  con- 
struct a  square  equivalent  to  the  sum  of 
three  squares,  whose  sides  are  m,  n,  and  ^a 
respectively. 

Draw  AB  equal  to  m. 

At  A  draw  AC  perpendicular  to  AB,  and  equal  to  n,  and 
draw  BC. 

At  C  draw  CD  perpendicular  to  BC,  and  equal  to  p,  and 
draw  BD. 

Then  the  square  described  with  its  side  equal  to  BD  will 
be  equivalent  to  the  sum  of  the  given  squares. 


B 


For, 


BD''  =  J5(7'  +  ^2  _  7n^  +  n?  -^p\  (§  273.) 


182 


PLANE   GEOMETRY.  — BOOK  IV. 


Proposition  XIII.     Problem. 

331.    To  construct  a  square  equivalent  to  the  difference  of 
two  given  squares. 


M 

B 

Its. 

N 

Let  M  and  iV  be  the  given  squares. 

To  construct  a  square  equivalent  to  the  difference  of  M 
and  N. 

Draw  the  indefinite  straight  line  AD. 

At  A  draw  AB  perpendicular  to  AD,  and  equal  to  a  side 
of  iV,  the  smaller  of  the  given  squares. 

With  ^  as  a  centre,  and  with  a  radius  equal  to  a  side  of 
M,  describe  an  arc  cutting  AD  at  C. 

Then  the  square  P,  described  with  its  side  equal  to  AC, 
will  be  equivalent  to  the  difference  of  M  and  iV! 

For,  AC^  =  BC^-  AB\  (§  274.) 

That  is,  area  P  =  area  M—  area  N.  (§  306.) 


Proposition  XIV.     Problem. 

332.    To  construct  a  square  equivalent  to  a  given  par- 
allelogram. 

K H 

C 


A    E 


Let  ABCD  be  the  given  parallelogram. 
To  construct  a  square  equivalent  to  ABCD, 


AREAS   OF  POLYGONS. 


183 


Draw  DE  perpendicular  to  AB,  and  construct  FG  a 
mean  proportional  between  AB  and  DE  (§  293). 

Then  the.  square  FGHK,  described  with  its  side  equal  to 
FGj  will  be  equivalent  to  ABCD. 

For  by  construction, 

AB:FG  =  FG:  DE. 

Whence,  FG^  =  AB  X  DE.  (§  231.) 

That  is,  area  FGHK  =  area  ABCD.    (§§  306,  310.) 

333.  CoR.  A  square  may  be  constructed  equivalent  to  a 
given  triangle  by  takirig  for  its  side  a  mean  proportional 
between  the  base  and  one-half  the  altitude  of  the  triangle. 

334.  ScH.  By  aid  of  §§  328  and  333,  a  square  may  be 
constructed  equivalent  to  a  given  polygon. 


Proposition  XV.    Problem. 

335.    With  a  given  straight  line  as  a  base,  to  construct  a 
rectangle  equivalent  to  a  given  rectangle. 


H 


O 


D 


B 


E 


F- 


Jjet  ABCD  be  the  given  rectangle,  and  EF  the  given  line. 
To  construct,  with  EF  as  a  base,  a  rectangle  equivalent 
to  ABCD. 

The  rectangle  EFGff,  constructed  with  EF  as  a  base, 
and  with  its  side  EH  equal  to  a  fourth  proportional  to  EFj 
AB,  and  AD  (§  291),  will  be  equivalent  to  ABCD. 

For  by  construction, 

EF:AB=AD:  EH. 
Whence,  EF  x  EH  =  AB  x  AD.  (§  231.) 

That  is,  area  EFGH  =  area  ABCD.  (§  305.) 


184 


PLANE  GEOMETRY. —BOOK  IV. 


Proposition  XVI.    Problem. 

336.    To  construct  a  rectangle  equivalent  to  a  given  square^ 
having  the  sum  of  its  base  and  altitude  equal  to  a  given  line. 


M 


C      D. 


iV 


E 


Let  M  be  the  given  square,  and  AB  the  given  line. 
To  construct  a  rectangle   equivalent  to   M,  having  the 
sum  of  its  base  and  altitude  equal  to  AB. 

With  AB  as  a  diameter,  describe  the  semi-circumference 
ADB. 

Draw  AC  perpendicular  to  AB,  and  equal  to  a  side  of  M. 

Draw  CF  parallel  to  AB,  intersecting  the  arc  ADB  at  D, 
and  draw  DE  perpendicular  to  AB. 

Then  the  rectangle  N,  constructed  with  its  base  and  alti- 
tude equal  to  BE  and  AE,  respectively,  will  be  equivalent 
toM. 

For,  AE.DE  =  DE:  BE.  (§  272,  I.) 

Whence,  'DEf  =  AE  x  BE.  (§  231.) 

That  is,  area  M  =  area  N.  (§§  306,  305.) 

Note.  The  above  construction  is  also  the  solution  of  the  fol- 
lowing problem : 

Given  the  sum  and  the  product  of  two  straight  lines,  to  construct 
the  lines. 


EXERCISES. 

56.  To  construct  a  triangle  equivalent  to  a  given  triangle,  having 
given  its  base. 

57.  To  construct  a  triangle  equivalent  to  a  given  square,  having 
given  its  base  and  an  angle  adjacent  to  the  base. 


AREAS  OF  POLYGONS. 


185 


Proposition  XVII.    Problem. 

337.  To  construct  a  rectangle  equivalent  to  a  given  square^ 
having  the  difference  of  its  base  and  altitude  equal  to  a  given 
line. 


A         '?         ]B 


N 


Let  M  be  the  given  square,  and  AB  the  given  line. 
To  construct  a  rectangle  equivalent  to  31,  having  the  dif- 
ference of  its  base  and  altitude  equal  to  AB. 

With  the  line  AB  as  a  diameter,  describe  the  circumfer- 
ence ADB. 

Draw  AC  perpendicular  to  AB,  and  equal  to  a  side 
of  M. 

Through  the  centre  0  draw  CO,  intersecting  the  circum- 
ference at  D  and  E. 

Then  the  rectangle  N,  constructed  with  its  base  and  alti- 
tude equal  to  CE  and  CD,  respectively,  will  be  equivalent 
toM. 

For,  CE-CD  =  DE  =  AB. 

That  is,  the  difference  of  the  base  and  altitude  of  N  is 
equal  to  AB. 

Again,  ^C  is  tangent  to  the  circle  at  A.  (§  169.) 

Whence,  CA^  =  CD  X  CE.  (§  282.) 

That  is,  area  M  =  area  N.  (§§  306,  305.) 

Note.  The  above  construction  is  also  the  solution  of  the  fol- 
lowing problem: 

Given  the  difference  and  the  product  of  two  straight  lines,  to 
construct  the  lines. 


186 


PLANE  GEOMETRY.  —  BOOK  IV. 


Proposition.  XVIII.     Problem. 

338.    To  construct  a  square  having  a  given  ratio  to  a  given 
square. 


m 

i/L.  1 N 

N 

M 

n 

A    m    D        -^        B 

Let  M  be  the  given  square,  and  let  the  given  ratio  be  that 
of  the  lines  m  and  n. 

To  construct  a  square  which  shall  have  to  M  the  ratio 
n:  m. 

On  the  straight  line  AB,  take  AD  =  m  and  DB  =  n. 

With  AB  as  a  diameter,  describe  the  semi-circumference 
ACB. 

Draw  DC  perpendicular  to  AB,  meeting  the  arc  ACB  at 
C,  and  draw  AC  and  BC. 

On  AC  take  CE  equal  to  a  side  of  Jf ;  and  draw  -E'i^ par- 
allel to  AB,  meeting  ^C  at  F. 

Then  the  square  N,  described  with  its  side  equal  to  CF, 
will  have  to  M  the  ratio  n  :  m. 


Por,  Z  ^C^  is  a  right  angle. 

Then  since  CD  is  perpendicular  to  AB, 


AC^      AD 

BC"      BD 

But  since  EF  is 

parallel  to  AB, 
CE      AC 
CF      BC 

Therefore, 

CE'       AC 
CF'       BC 

That  is, 

area  M      m 
area  N       n 

m 


m 


(§  196.) 
(§  271.) 

(§245.) 
(§  306.) 


AREAS   OF  POLYGONS. 


187 


Proposition  XIX.     Problem. 

339.    To  construct  a  polygon  similar  to  a  given  polygon, 
and  having  a  given  ratio  to  it. 


A  B 


A! B' 


Let  A-E  be  the  given  polygon,  and  let  the  given  ratio 
be  that  of  the  lines  m  and  n. 

To  construct  a  polygon  similar  to  A-E^  and  having  to 
it  the  ratio  n  :  m. 

Construct  AB ',  the  side  of  a  square  which  shall  have  to 
the  square  described  upon  AB  the  ratio  w  :  m  (§  338). 

Upon  the  side  A'B',  homologous  to  AB,  construct  the 
polygon  A'-E'  similar  to  A-E  (§  298). 

Then  A-E'  will  have  to  A-E  the  ratio  n :  m. 

For  since  A-E  is  similar  to  A'-E', 
A-E         AB'' 


A'-E'      A'B 


72 


But  by  construction, 


Whence, 


Ml 

-AW' 

A-E 

A-E' 


m 


m 
n 


(§  323.) 


EXERCISES. 

58.  To  construct  a  triangle  equivalent  to  a  given  square,  having 
given  its  base  and  the  median  drawn  from  the  vertex  to  the  base. 

59.  To  construct  a  square  equivalent  to  twice  a  given  square. 


isa 


PLANE   GEOMETRY.  —  BOOK  IV. 


Proposition   XX.     Problem. 

340.    To  construct  a  polygon  similar  to  one  of  two  given 
polygons^  and  equivalent  to  the  other. 


Let  M  and  N  be  the  given  polygons. 

To   construct   a   polygon   similar  to  Jf,  and   equivalent 

toiv: 

Let  AB  be  any  side  of  M. 

Construct  m,  the  side  of  a  square  equivalent  to  Jf,  and  n, 
the  side  of  a  square  equivalent  to  N  (§  334). 

Construct  A'B'  a  fourth  proportional  to  m,  n,  and  AB 
(§  291). 

Upon  the  side  A'B',  homologous  to  AB,  construct  the 
polygon  P  similar  to  M  (§  298). 

Then  P  will  be  equivalent  to  K 

Por  since  M  is  similar  to  P, 


area  M       AB ' 


area  P      A'B'^ 

But  by  construction,  we  have 

AB        m 

AB'  ~  n  ' 

Whence, 

area  M       m^ 
area  P        n''  ' 

But, 

m^  =  area  M, 

and 

n^  =  area  N. 

Whence, 

area  M      area  M 
area  P      area  N 

Therefore, 

area  P  =  area  N. 

(§  323.) 


(§  306.) 


AREAS   OF  POLYGONS.  189 


EXERCISES. 


60.  To  construct  an  isosceles  triangle  equivalent  to  a  given  tri- 
angle, having  its  base  coincident  with  a  side  of  the  given  triangle. 

61.  To  construct  a  riiombus  equivalent  to  a  given  parallelogram, 
having  one  of  its  diagonals  coincident  with  a  diagonal  of  the 
parallelogram. 

62.  To  construct  a  triangle  equivalent  to  a  given  triangle,  liaving 
given  two  of  its  sides. 

63.  To  construct  a  right  triangle  equivalent  to  a  given  square, 
having  given  its  hypotenuse. 

64.  To  construct  a  right  triangle  equivalent  to  a  given  triangle, 
having  given  its  hyi^otenuse. 

65.  To  construct  an  isosceles  triangle  equivalent  to  a  given  tri- 
angle, having  given  one  of  its  equal  sides.  How  many  different 
triangles  can  be  constructed  ? 

66.  To  draw  a  line  parallel  to  the  base  of  a  triangle  dividing  it 
into  two  equivalent  parts.     (§320.) 

67.  To  draw  through  a  given  point  within  a  parallelogram  a 
straight  line  dividing  it  into  two  equivalent  parts. 

68.  To  construct  a  parallelogram  equivalent  to  a  given  trapezoid, 
having  a  side  and  two  adjacent  angles  equal  to  one  of  the  non-paral- 
lel sides  and  the  adjacent  angles  of  the  trapezoid. 

69.  To  draw  through  a  given  point  in  a  side  of  a  parallelogram 
a  straight  line  dividing  it  into  two  equivalent  parts. 

70.  To  draw  a  straight  line  perpendicular  to  the  bases  of  a  trape- 
zoid, dividing  the  trapezoid  into  two  equivalent  parts. 

(Draw  a  line  connecting  the  middle  points  of  the  bases.) 

71.  To  draw  through  a  given  point  in  the  smaller  base  of  a  trape- 
zoid a  straight  line  dividing  the  trapezoid  into  two  equivalent  parts. 

72.  To  construct  a  triangle  similar  to  two  given  similar  triangles, 
and  equivalent  to  their  sum. 

73.  To  construct  a  triangle  similar  to  two  given  similar  triangles, 
and  equivalent  to  their  difference. 


Book  Y. 


REGULAR     POLYGONS.  —  MEASUREMENT     OP 
THE    CIRCLE. 


341.    Def.    a   Tegular   'polygon   is   a   polygon   which   is 
both  equilateral  and  equiangular. 


Proposition   L     Theorem. 

342.   A  circle  can  he  circumscribed  ahouty  or  inscribed  in, 
any  regular  polygon. 


Let  ABODE  hQ  a  regular  polygon. 

I.   To  prove  that  a  circle   can   be   circumscribed   about 
ABODE. 

Let  a  circumference  be  described  through  the  vertices  A, 
B,  and  G  (§  223). 

Let  0  be  the  centre  of  the  circumference,  and  draw  OA, 
OB,  00,  and  OD. 
Then  since  ABODE  is  equiangular, 

Z  ABO  =  Z  BOD. 
190 


REGULAR  POLYGONS.  191 

And  since  the  triangle  OBC  is  isosceles, 

Z  OBC  =  Z  OCB.  (§  91.) 

Then,   Z  ABC-  Z  OBC  =  Z  BCD  -  Z  OCB. 

That  is,  Z  OB  A  =  Z  06'i>. 

Also,  OB  =  OC.  (§  143.) 

And  since  ABCDE  is  equilateral,  • 
AB  =  CD. 

Therefore,  A  OAB  =  A  OCD.  (§  63.) 

Whence,  OA  =  OD.  (§  66.) 

Then  the  circumference  passing  through  A,  i?,  and  C 
also  passes  through  D. 

In  like  manner,  it  may  be  proved  that  the  circumference 
passing  through  Bj  C,  and  D  also  passes  through  E. 

Hence,  a  circle  can  be  circumscribed  about  ABCDE. 

II.   To  prove  that  a  circle  can  be  inscribed  in  ABCDE. 

Since  AB,  BC,  CD,  etc.,  are  equal  chords  of  the  circum- 
scribed circle,  they  are  equally  distant  from  0.  (§  164.) 

Hence,  a  circle  described  with  0  as  a  centre,  and  with  the 
perpendicular  OF  from  0  to  any  side  AB  as  a  radius,  will 
be  inscribed  in  ABCDE. 

343.  Def.  The  centre  of  a  regular  polygon  is  the  common 
centre  of  the  circumscribed  and  inscribed  circles. 

The  anr/le  at  the  centre  is  the  angle  between  the  radii 
drawn  to  the  extremities  of  any  side;  as  A  OB. 

The  radius  is  the  radius  of  the  circumscribed  circle ;  as 
OA. 

The  apothem  is  the  radius  of  the  inscribed  circle ;  as  OF. 

344.  CoR.  From  the  equal  triangles  OAB,  OBC,  etc.,  we 
have 

Z  AOB  =  Z  BOC  =  Z  COD,  etc.  (§  66.) 

Then  each  of  these  angles  is  equal  to  four  right  angles 
divided  by  the  number  of  sides  of  the  polygon.  (§  37.) 

That  is,  the  angle  at  the  centre  of  a  regular  polygon  is  equal 
to  four  right  angles,  divided  by  the  number  of  sides. 


192  PLANE   GEOMETRY.  —  BOOK  V. 

Proposition  II.     Theorem. 

345.    If  the  circumference  of  a  circle  he  divided  into  any 
number  of  equal  arcs, 

I.    Their  chords  form  a  regular  inscribed  polygon. 

II.    Tangents  at  the  points  of  division  forin  a  regular  cir- 
cumscribed polygon. 


Let  the  circumference  ACD  be  divided  into  any  number 
of  equal  arcs,  AB,  BC,  CD,  etc. 

I.  To  prove  ABODE  a  regular  polygon. 

Now,     chord  AB  =  chord  BC=  chord  CD,  etc.     (§  158.) 
Again,  since  arc  AB  =  arc  BC=  arc  CD,  etc.,  we  have 

arc  BCDE  =  arc  CD£JA  =  arc  DEAD,  etc. 
Whence,     Z  EAB  =ZABC=ZBCD,  etc.  (§  193.) 

Therefore,  the  polygon  ABCDE  is  regular.  (§  341.) 

II.  Let  FGHKL  be  a  polygon  whose  sides  LF,  EG,  etc., 
are  tangent  to  the  circle  at  the  points  A,  B,  etc.,  respectively. 

To  prove  FGHKL  a  regular  polygon. 

In  the  triangles  ABE,  BCG,  CDH,  etc.,  we  have 

AB  =  BC=  CD,  etc. 

Also,  since  arc  AB  =  arc  BC  =  arc  CD,  etc.,  we  have 

Z  BAF  =  Z  ABE  =  Z  CBG  =  Z  BCG,  etc. 

(§  197.) 

Hence,  the  triangles  ABE^  BCG,  etc.,  are  all  equal  and 

isosceles.  (§§  68,  94.) 


•  REGULAR  POLYGONS.  193 

Therefore,  ZF  =  Z.G=Z.H,  etc., 

and  BF=:BG==CG  =  CH,  etc.  (§  66.) 

Whence,  FG  =  GH  =  HK,  etc. 

Therefore,  the  polygon  FGHKL  is  regular.  (§  341.) 

Proposition  III.     Theorem. 

346.  Tangents  to  a  circle  at  the  middle  points  of  the  arcs 
subtended  by  the  sides  of  a  regular  inscribed  polygon,  form 
a  regular  circumscribed  polygon. 


Let  ABODE  be  a  regular  polygon  inscribed  in  the  circle 
AC. 

Let  A'B'C'D'E'  be  a  polygon  whose  sides  A'B',  B'C, 
etc.,  are  tangent  to  the  circle  at  the  middle  points  F,  G, 
etc.,  of  the  arcs  AB,  BC,  etc. 

To  prove  AB'C'B'E'  a  regular  polygon. 

We  have,    arc  AB  =  arc  BC  =  sltq  CD,  etc.         (§  157.) 
Whence,     arc  AF  =  arc  BF  =  arc  BG  =  arc  CG,  etc. 
Therefore,  arc  FG  =  arc  Gil  =  arc  HK,  etc. 
Whence,  the  polygon  A'B'C'D'E'  is  regular.    (§  345,  II.) 

347.    Cor.   Let  0  be  the  centre  of  the  circle,  and  draw 
OF,  OL,  and  OA'. 
Then,  OA'  bisects  Z  FOL.  (§  175.) 

Whence,  OA'  passes  through  A.  (§  154.) 

That  is,  the  radii  of  a  regular  circumscribed  polygon  in- 
tersect the  circumference  in  points  which  are  the  vertices  of  a 
regular  inscribed  polygon  having  the  same  number  of  sides. 


194 


PLANE  GEOMETKY.  —  BOOK  V. 


Proposition  IV.     Theorem. 


348.   Regular  polygons  of  the  same  number  of  sides  are 
similar. 


A  B  A  B' 

Let  A-E  and  A!-E'  be  two  regular  polygons  of  the  same 
number  of  sides. 

To  prove  A-E  and  Al-E'  similar. 

The  sum  of  all  the  angles  of  A-E  is  equal  to  the  sum  of 
all  the  angles  of  A!-E'.  (§  126.) 

Whence,  each  angle  of  A-E  equals  each  angle  of  A'-E'. 
Again,  since  AB  =  BC,  etc.,  and  A'B'  =  B' C ,  etc., 


AB 

A'B' 


BC 


B'C       CD 


^^      etc 


Therefore,  A-E  and  A'-E'  are  similar. 


(§  252.) 


Proposition  V.     Theorem. 


349.  The  perimeters  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  their  radii,  or  as  their 
apothems. 


Wb' 


Let  A-E  and  A'-E'  be  two  regular  polygons  of  the  same 
number  of  sides. 


REGULAR  POLYGONS.  195 

Let  P  and  F'  denote  their  perimeters,  R  and  R'  theit 

radii,  and  r  and  r'  their  apothems. 

rr  P         R         r 

To  prove  _  =  _  =  _. 

Let  0  and  0'  be  the  centres  of  the  polygons  A-E  and 
A'-E'. 

Draw  OJ,  OB,  O'A',  and  O'l?'. 

Also,  draw  OE  and  O'i'"'  perpendicular  to  AB  and  /l'^', 
respectively. 

Then,  OA  =  72,  0'^'  =  R',  OF  =  r,  and  O'i^'  =  r'. 
Now  in  the  isosceles  triangles  OAB  and  O'A'B', 

AAOB=^  A  A'O'B'.  .(§  344.) 

And  since  OA  =  Oi?,  and  O'A!  =  O'i?',  we  have 
OA   ^    OB 
O'A'       O'B'' 
Therefore,  the  triangles  OAB  and  O'A'B'  are  similar. 

(§  260.) 
Whence,  _g.  =  |-  =  ^.      (§§  253,  IL,  263.) 

But  the  polygons  A-E  and  A'-E'  are  similar.        (§  348.) 

Whence,  :F  =  if7-  (§268.) 

Therefore, 


350.    CoR.     Let  K  and  7f'  denote  the  areas  of  the  poly- 
gons A-E  and  A'-E'. 

Then,  #7  =  44-  (§320.) 


K' 

IB' 

-2 

AB 
A'B' 

R 
R' 

r 

''  r' 

K  _ 
K'  ~ 

R^ 

7-2 

But, 

Whence, 

That  is,  the  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  s<iuares  of  their 
radii,  or  as  the  squares  of  their  apotheiiis. 


196 


PLANE   GEOMETEY.  —  BOOK  Y. 


Proposition  VI.     Theorem. 

351.    The  area  of  a  regular  polygon  is  equal  to  one-half 
the  product  of  its  perimeter  and  apothem. 


A       F      D 


Let  r  denote  the  apothem  OF,  and  P  the  perimeter,  of 
the  regular  polygon  A-U. 

To  prove  area  A-E  =  ^  P  X  r. 

Draw  the  radii  OA,  OB,  OC,  etc.,  forming- a  series  of  tri- 
angles, OAB,  OBC,  etc.,  whose  common  altitude  is  r. 
Now,  area  OAB  =  ^  AB  x  r, 

area  OBC  =  ^BC  X  r,  etc.  (§  313.) 

Adding,  we  have 

area  A-E  =  \  (AB  +  j5C  +  etc.)  X  r 
=  ^P  Xr. 


Proposition  VII.     Problem. 
352.    To  inscribe  a  square  in  a  given  circle. 

B 


Let  AC  hQ  the  given  circle. 


REGULAR  POLYGONS.  197 

To  inscribe  a  square  in  AC. 

Draw  the  diameters  AC  and  BI)  perpendicular  to  each 
other,  and  draw  AB,  BC,  CD,  and  DA. 
Then  ABCD  is  an  inscribed  square. 
For,  arc  AB  =  arc  BC  =  arc  CD  =  arc  DA.  (§  191.) 

Whence,  ABCD  is  an  inscribed  square.  (§  345,  I.) 

353.  CoK.     Denoting  the  radius  OA  by  Rj  we  have 

AB'=0T+  OTf_=  2  B\  (§  273.) 

Whence,  AB  =  R  V2. 

That  is,  the  side  of  an  inscribed  square  is  eriual  to  the 
radius  of  the  circle  multiplied  by  V2. 

Proposition  VIIL     Problem? 

354.  To  ifiscribe  a  regular  hexagon  in  a  given  circle. 


Let  AC  be  the  given  circle. 

To  inscribe  a  regular  hexagon  in  AC. 

Draw  any  radius  OA. 

With  ^  as  a  centre,  and  OA  as  a  radius,  describe  an  arc 
cutting  the  given  circumference  at  B,  and  draw  AB. 

Then  AB  is  a  side  of  a  regular  inscribed  hexagon. 

For  drawing  OB,  the  triangle  OAB  is  equilateral. 

Whence,  Z  AOB  is  one-third  of  two  right  angles.  (§  82.) 

Then,  /.  AOB  is  one-sixth  of  four  right  angles,  and  AB  is 
a  side  of  a  regular  inscribed  hexagon.  (§  345,  I.) 

Hence,  to  inscribe  a  regular  hexagon  in  a  circle,  apply 
the  radius  six  times  as  a  chord. 


198 


PLANE  GEOMETKY.  —  BOOK  V. 


355.  Cor.  I.     The  side  of  a  regular  inscribed  hexagon  is 
equal  to  the  radius  of  the  circle. 

356.  Cor.  II.    By  joining   the   alternate  vertices  of  the 
hexagon,  there  is  formed  an  inscribed  equilateral  triangle. 

357.  Cor.  III.  Let  AB  be  a  side 
of  a  regular  hexagon,  inscribed  in  the 
circle  AD  whose  radius  is  B. 

Draw  the  diameter  BC,  and  join  ^C. 

Then  ^C  is  a  side  of  an  inscribed 
equilateral  triangle. 

Now  ABC  is  a  right  triangle. 

Then,      Aa^=^BC^  -  AB" 
=  (2By-B' 

Whence,  AC  =  B  V3. 

That  is,  the  side  of  an  inscribed  equilateral   triangle   is 
equal  to  the  radius  of  the  circle  multiplied  by  V3. 


(§  196.) 
(§  274.) 


4.B^-B^  =  3  B^ 


Proposition  IX.     Problem. 
358.    To  inscribe  a  regular  decagon  in  a  given  circle. 


Let  AC  hQ  the  given  circle. 

To  inscribe  a  regular  decagon  in  AC. 

Draw  any  radius  OA. 

Divide  OA  in  extreme  and  mean  ratio  (§  296),  so  that 

OA  :  DM  =  DM '.  AM.  (1) 


KEGULAR  POLYGONS.  190 

With  ^  as  a  centre,  and  OM  as  a  radius,  describe  an  arc 
cutting  the  given  circumference  at  B,  and  draw  AB. 
Then  AB  is  a,  side  of  a  regular  inscribed  decagon. 
For  draw  OB  and  BM. 

Then  in  the  triangles  OAB  and  ABM,  Z  A  is  common. 
And  since  OM  =  AB,  the  proportion  ( 1 )  gives 

OA-.AB  =  AB  :  AM. 
Therefore,  OAB  and  ABM  are  similar.  (§  260.) 

Whence,  Z  ABM  =  ZAOB.  ( 2 ) 

Now  the  triangle  OAB  is  isosceles. 
Hence,  the  similar  triangle  ABM  is  isosceles,  and 

AB^BM=OM. 
Therefore,  Z  OBM  =  Z  AOB.  ( 3 ) 

Adding  (2)  and  (3),  we  liave 

Z0BA  =  2ZA0B,  (4) 

But  since  the  triangle  OAB  is  isosceles, 

2  Z  OB  A  +  ZAOB  =  180°.  (§  82.) 

Then  by  (4), 

5ZA0B  =  180°,  or  Z  AOB  =  36°. 

Therefore,  ZAOB  is  one-tenth  of  four  right  angles,  and 

AB  is  a  side  of  a  regular  inscribed  decagon.  (§  345,  I.) 

Hence,  to  inscribe  a  regular  decagon  in  a  circle,  divide 

the  radius  in  extreme  and  mean  ratio,  and  apply  the  greater 

segment  ten  times  as  a  chord. 

359.  CoR.  I.    B]/  joining   the   alternate  vertices   of   the 
decagon,  there  is  formed  a  regular  inscribed  'pentagon. 

360.  CoR.  II.  Denoting  the  radius  of  the  circle  by  U,  we 
have 

AB=OM=  ^  ^^f  ~  ^)  .  (§  297.) 

This  is  an  expression  for  the  side  of  a  regular  inscribed 
decagon  in  terms  of  the  radius  of  the  circle. 


Ex.  1.   The  angle  at  the  centre  of  a  regular  polygon  is  the  supple- 
ment of  the  angle  of  the  polygon. 


200  PLAKE   GEOMETRY.  —  BOOK  V. 

Proposition  X.     Problem! 

361.    To  construct  the  side  of  a  regular  pentedecagon 
scribed  in  a  given  circle. 

^       C 


Let  MN  be  an  arc  of  the  given  circumference. 

To  construct  the  side  of  a  regular  inscribed  polygon  of 
fifteen  sides. 

Construct  AB  equal  to  a  side  of  a  regular  inscribed  hexa- 
gon (§  354),  and  AC  equal  to  a  side  of  a  regular  inscribed 
decagon  (§  358),  and  draw  BC. 

Then  ^C  is  a  side  of  a  regular  inscribed  pentedecagon. 

Por  the  arc  BC  is  ^  —  ^,  or  ^,  of  the  circumference. 

Hence,  the  chord  ^C  is  a  side  of  a  regular  inscribed 
pentedecagon.  (§  345,  I.) 

362.  ScH.  By  bisecting  the  arcs  AB,  BC,  etc.,  in  the 
figure  of  Prop.  VII.,  we  may  construct  a  regular  inscribed 
octagon  (§  345,  I.) ;  and  by  continuing  the  bisection,  we 
may  construct  regular  inscribed  polygons  of  16,  32,  64,  etc., 
sides. 

In  like  manner,  by  aid  of  Props.  VIII. ,  IX.,  and  X.,  we 
may  construct  regular  inscribed  polygons  of  12,  24,  48,  etc., 
or  of  20,  40,  80,  etc.,  or  of  30,  60,  120,  etc.,  sides. 

By  drawing  tangents  to  the  circumference  at  the  vertices 
of  any  one  of  the  above  inscribed  polygons,  we  may  con- 
struct a  regular  circumscribed  polygon  of  the  same  number 
of  sides.  .  (§  345,  II.) 

EXERCISES. 

2.  An  equilateral  polygon  inscribed  in  a  circle  is  regular. 

3.  An  equiangular  polygon  circumscribed  about  a  circle  is  regular. 


HEGtJLAR  POLYGONS.  201 

Find  the  angle,  and  the  angle  at  the  centre, 

4.  Of  a  regular  pentagon. 

5.  Of  a  regular  dodecagon. 

6.  Of  a  regular  polygon  of  32  sides. 

7.  Of  a  regular  polygon  of  25  sides. 

If  r  represents  the  radius,  a  the  apothera,  s  the  side,  and  k  the 
area,  prove  that: 

8.  In  an  equilateral  triangle,  a=irj  and  k  =  if^  Vs. 

9.  In  a  square,  a  =  \r  V2,  and  A;  =  2  r^. 

10.  In  a  regular  hexagon,  a=\r  V3,  and  fc  =  |  r^  V3. 

11.  In  an  equilateral  triangle,  r  =  2a,  s  =  2a  VS,  and  fc  =  3  a'^  VS. 

12.  In  a  square,  r  =  a  V2,  s  =  2  a,  and  A;  =  4  a^. 

13.  In  a  regular  hexagon,  r  =  ia  V3,  and  A;  =  2  a'-^  Vs. 

14.  In  an  equilateral  triangle,  express  r,  a,  and  k  in  terms  of  s. 

15.  In  a  square,  express  r,  a,  and  A:  in  terms  of  a. 

16.  In  a  regular  hexagon,  express  a  and  A:  in  terms  of  s. 

17.  In  an  equilateral  triangle,  express  r,  a,  and  »  in  terms  of  A;. 

18.  In  a  square,  express  r,  a,  and  s  in  terms  of  fc. 

19.  In  a  regular  hexagon,  express  r  and  a  in  terms  of  A:. 

20.  The  apothem  of  an  equilateral  triangle  is  one-third  the  alti- 
tude of  the  triangle. 

21.  The  sides  of  a  regular  polygon  circumscribed  about  a  circle 
are  bisected  at  the  points  of  contact. 

22.  The  radius  drawn  from  tlie  centre  of  a  regular  polygon  to  any 
vertex  bisects  the  angle  at  that  vertex. 

23.  The  diagonals  of  a  regular  pentagon  are  equal. 

24.  The  figure  bounded  by  the  five  diagonals  of  a  regular  penta- 
gon is  a  regular  pentagon. 

25.  The  area  of  a  regular  inscribed  hexagon  is  a  mean  propor- 
tional between  the  areas  of  an  inscribed,  and  of  a  circumscribed 
equilateral  triangle. 

26.  If  the  diagonals  AC  and  BE  of  the  regular  pentagon 
ABODE  intersect  at  F,  prove  AC  =  AE  +  BF.     (Ex.  23. ) 

27.  In  the  figure  of  Prop.  IX.,  prove  that  OM  is  the  side  of  a 
regular  pentagon  inscribed  in  a  circle  which  is  circumscribed  about 
the  triangle  OBM. 


202 


PLANE   GEOMETRY.  —  BOOK   V 


Proposition  XI.     Theorem. 


TH^i-R^ 


363.  If  d  regular  polygon  be  inscribed  in,  or  circumscribed 
about,  a  circle,  and  the  number  of  its  sides  be  indefinitely 
increased, 

I.    Its  perimeter  approaches  the  circumference  as  a  limit. 

II.   Its  area  approaches  the  area  of  the  circle  as  a  limit. 


Let  p  and  P  denote  the  perimeters,  and  k  and  K  the 
areas,  of  two  regular  polygons  of  the  same  number  of  sides, 
respectively  inscribed  in,  and  circumscribed  about,  a  circle. 

Let  C  denote  the  circumference  of  the  circle,  and  S  its 
area. 

I.  To  prove  that  if  the  number  of  sides  of  the  polygons 
be  indefinitely  increased,  P  and  p  approach  the  limit  C. 

Let  A'B'  be  a  side  of  the  polygon  whose  perimeter  is  P. 

Draw  the  radius  OF  to  its  point  of  contact ;  also,  draw 
OA'  and  OB'  cutting  the  circumference  at  A  and  B,  and 
draw  AB. 

Then,  AB  is   a   side   of  the  polygon  whose   perimeter 


is  p. 


Now  the  two  polygons  are  similar. 

Whence, 

P        OA' 
V        OF 

Then, 

P-.p  _0A'  -OF 
P                 OF 

Or, 

P-P  =  ^><{0A'-C 

(§  347.) 
(§  348.) 

(§  349.) 

(§  237.) 

(1) 


REGULAR  POLYGONS.  203 

Now  let  the  number  of  sides  of  each  polygon  be  indefi- 
nitely increased,  the  two  polygons  continuing  to  have  the 
same  number  of  sides. 

Then  the  length  of  each  side  will  be  indefinitely  dimin- 
ished, and  A!F  will  approach  the  limit  0. 

Therefore,  OA!—  OF  will  approach  the  limit  0. 

Whence,  by  ( 1 ),  P—  p  will  approach  the  limit  0 ;  that  is, 
the  difference  between  the  perimeters  of  the  polygons  will 
approach  the  limit  0. 

But  the  circumference  of  the  circle  is  evidently  less  than 
the  perimeter  of  tlie  circumscribed  polygon;  and  it  is  greater 
than  the  perimeter  of  tlie  inscribed  polygon.  (Ax.  6.) 

Hence,  the  difference  between  the  perimeter  of  either 
polygon  and  the  circumference  of  the  circle  will  approach 
the  limit  0. 

That  is,  P  —  C  and  C  —  p  will  each  approach  the  limit  0. 

Therefore,  P  and  p  will  each  approach  the  limit  C. 

II.   To  prove  that  K  and  k  approach  the  limit  S. 

Since  the  polygons  whose  sides  are  AB  and  A'B'  are 

similar, 

"  '^  (§350.) 

Whence,         ^^^  =  "^  ^r'    =  =F '  (§  ^74.) 

That  is, 

Now  let  the  number  of  sides  of  each  polygon  be  indefi- 
nitely increased,  the  two  polygons  continuing  to  have  the 
same  number  of  sides. 

Then  A'F,  and  therefore  K—k,  will  approach  the  limit  0. 

But  the  area  of  the  circle  is  evidently  less  than  the  area 
of  the  circumscribed  polygon,  and  greater  than  the  area  of 
the  inscribed  polygon. 

Hence,  K — S  and  S  —  k  will  each  approach  the  limit  0. 

Therefore,  K  and  k  will  each  approach  the  limit  S. 


K 

k  ' 

OA'' 
OF'' 

K-k 

k 

_0J 

[''-OF' 
OF' 

A'F' 
OF' 

K- 

-k  = 

OF' 

204  PLANE   GEOMETRY.  —  BOOK  V. 

364.  Cor.  1.  If  a  regular  2^olygon  he  inscrihecl  in  a  circle, 
and  the  number  of  its  sides  be  indefinitely  increased,  its  apo- 
them  approaches  the  radium  of  the  circle  as  a  limit. 

2.  If  a  regular  polygon  be  circumscribed  about  a  circle,  and 
the  number  of  its  sides  be  indefinitely  increased,  its  radius 
approaches  the  radius  of  the  circle  as  a  limit. 

MEASUREMENT   OF    THE   CIRCLE. 
Proposition  XII.     Theorem. 

365.  The  circumferences  of  two  circles  are  to  each  other  as 
their  radii.  A    , 


Let  Cand  C  denote  the  circumferences,  and  B  and  B' 
the  radii,  of  two  circles. 

m  C  B 

To  prove  _  =  -. 

Inscribe  in  the  circles  regular  polygons  having  the  same 
number  of  sides,  and  let  P  and  P'  denote  their  perimeters. 

Then,  §-'  =  W'  ^^^^^'^ 

Whence,  P  xB'  =  P'xB.  (§  231.) 

Now  let  the  number  of  sides  of  each  polygon  be  indefi- 
nitely increased,  the  two"  polygons  continuing  to  have  the 
same  number  of  sides. 

Then,  P  X  B'  will  approach  the  limit  C  X  B', 
and         P'  XB  will  approach  tlie  limit  6"  X  B.  (§  363,  I.) 


MEASUREMENT  OF  THE  CIRCLE.  205 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 
Whence,      C  X  J?'  =  C"  X  ^,  or  ^,  =  :|^ .  (§  233.) 

366.  Cor.  I.  Multiplying  the  terms  of  the  ratio  — -  by 
2,  we  have  ^       2  ^ 

Or,  denoting  the  diameters  of  the  circles  by  D  and  Z)', 

CD'' 

That  is,  the  circumferences  of  two  circles  are  to  each  other 
as  their  diameters. 

367.  Coii.  II.   The  proportion 

g,  =  |-,  may  be  written  f  =  f^  •         (§  234.) 

That  is,  the  ratio  of  the  circumference  of  a  circle  to  its 
diameter  has  the  same  value  for  every  circle. 

This  constant  value  is  denoted  by  the  symbol  ir. 

That  is,  J)^'''  (^^ 

It  is  shown  by  the  methods  of  higher  mathematics  that 
the  ratio  ir  is  incommensurable  ;  hence,  its  numerical  value 
can  only  be  obtained  approximately. 

Its  value  to  the  nearest  fourth  decimal  place  is  3.1416. 

368.  CoR.  III.   The  equation  (1)  of  §  367  gives 

C  =  ttD. 

That  is,  the  circumference  of  a  circle  is  equal  to  its  diame- 
ter midtixjlied  by  tt. 

We  also  have  (7=2  ttB. 

That  is,  the  circumference  of  a  circle  is- equal  to  its  radius 
multiplied  by  2  tt. 

369.  Def.  In  circles  of.  different  radii,  similar  arcs, 
similar  segments,  and  similar  sectors  are  those  which  corre- 
spond to  equal  central  angles. 

•of  the  . 
UNIVERSITY 


206  PLANE   GEOMETRY.  —  BOOK  V. 


Proposition   XIII.     Theorem. 

370.    The  area  of  a  circle  is  equal  to  one-half  the  product 
of  its  circumference  and  radius. 


)  u^.^^y-  -^ 


Let  B  denote  the  radius,  C  the  circumference,  and  S  the 
area,  of  a  circle. 

To  prove  ^      8=\CxB. 

Circumscribe  about  the  circle  a  regular  polygon. 

Let  P  denote  its  perimeter,  and  K  its  area. 

Then  since  the  apothem  of  the  polygon  is  J?,  we  have 

K=\PxB.  (§351.) 

Now  let  the  number  of  sides  of  the  polygon  be  indefinitely 
increased. 

Then,  K  will  approach  the  limit  S, 

and         \  P  xR  will  approach  the  limit  ^  C  X  H.  (§  363.) 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 

Whence,  S  =  ^  C  X  B. 

371.    Cor.  I.   We  have  C  =  2  ttM.  (§  368.) 

Whence,  S  =X'r  R  X  j[  R  =  tt  R^- 

That  is,  the  area  of  a  circle  is  equal  to  the  square  of  its 
radius  multiplied  hy  tt. 

Again,  ;S  =  J  tt  X  4  JS^  _  i  tt  X  (2Ry. 

Or,  denoting  the  diameter  of  the  circle  by  D, 

That  is,  the  area  of  a  circle  is  equal  to  the  square  of  its 
diameter  multiplied  hy  ^ir. 


2  - 


MEASUREMENT   OF  THE  CIRCLE.  207 

372.    Cor.  II.   Let   S  and  S'  denote  the  areas  of  two 
circles,  B  and  B'  tlieir  radii,  and  D  and  B'  their  diameters. 
S         trR'         R^ 


Then, 


S'      ttR"'      R' 


That  is,  ^Ae  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii,  or  as  the  squares  of  their  diameters. 

373.  Cor.  III.  A  sector  is  the  same  part  of  the  circle 
that  its  arc  is  of  the  circumference. 

Hence,  denoting  the  area  and  arc  of  the  sector  by  s  and  c, 
and  the  area  and  circumference  of  the  circle  by  S  and  C, 
we  have  s        c  ^,  S 

But,  7,  =  h^-  (§  370.) 

Whence,  s  =  }^c  X  R- 

That  is,  the  area  of  a  sector  is  equal  to  one-half  the  jprod-l 

uct  of  its  arc  and  radius.  » 

374.  CoR.  IV.  Since  similar  sectors  are  like  parts  of  the 
circles  to  which  they  belong  (§  369),  it  follows  that 

Similar  sectors  are  to  each  other  as  the  squares  of  their 
radii. 

EXERCISES. 

28.  The  area  of  a  circle  is  equal  to  four  times  the  area  of  the 
circle  described  upon  its  radius  as  a  diameter. 

29.  The  area  of  one  circle  is  2|  times  the  area  of  another.  If  the 
radius  of  the  first  is  15,  what  is  the  radius  of  the  second  ? 

30.  The  diameters  of  two  circles  are  64  and  88,  respectively. 
What  is  the  ratio  of  their  areas? 

31.  The  radii  of  three  circles  are  3,  4,  and  12,  respectively.  What 
is  the  radius  of  a  circle  equivalent  to  their  sum  ? 

32.  Find  the  radius  of  a  circle  whose  area  is  one-half  the  area  of 
a  circle  whose  radius  is  9. 


208  PLANE   GEOMETRY.  —  BOOK  V. 


Proposition  XIV.     Problem. 

375.  Given  p  and  P,  the  perimeters  of  a  regular  inscribed 
and  of  a  regular  circumseribed  polygon  of  the  same  number 
of  sides,  to  find  p'  and  P',  the  j^eriTnetei's  of  the  regular  in- 
scribed and  chrymnscribed  polygons  having  double  the  number 
of  sides.  i 


a!         M  F     ■  N  B' 


I 


^^7      ^ 


O 

Let  AB  be  a  side  of  the  polygon  whose  perimeter  is  p. 

Draw  the  radius  OF  to  the  middle  point  of  the  arc  AB ; 
also,  draw  OA  and  OB  cutting  the  tangent  at  F  at  A' 
and  B'. 

Then,  A'B^  is  a  side  of  the  i)olygoii  whose  perimeter 
is  P.  (§§346,347.) 

Draw  AF  and  BF;  also,  draw  AM  and  BN  tangent  to 
the  circle  at  A  and  B,  meeting  A'B'  at  M  and  iV". 

Then  AF  and  IIH  are  sides  of  the  polygons  whose  per- 
imeters are  p'  and  P\  (§  345.) 

Hence,  if  n  denotes  the  number  of  sides  of  the  polygons 
whose  perimeters  are  p  and  P,  and  therefore  2  7i  tlie  num- 
ber of  sides  of  the  polygons  whose  perimeters  are  p'  and  P', 
we  have 

AB  =  P,  A'B'  =  ?.,AF=  -f^,  and  Jim  =  ^.(1) 
n'  n  2n  2n    ^    ^ 

Draw  OM)  then  OM  bisects  ZA'OF.  (§  175.) 

Whence,  ^^  =  -2^.  n  249.) 

'  MF        OF  ^s^^^-; 

But  OA^  and  OF  are  the  radii  of  the  polygons  whose  per- 
imeters are  P  and  p. 


MEASUKEMENT  OF  THE   CIRCLE.  209 

(§  349.) 

(§  236.) 


P        OA' 

P       'OF  ' 

Then, 

P       A'M 

P       MF  ' 

Therefore, 

P-\-p  _A'M^MF 
p                 MF 

A'F       i  A'B' 
MF       i  MN 
P 

Then  by  (1 

), 

P-\-p        2n        2P 
p             P'         P'' 

An 

Clearing  of  fractions, 

P'(P+V)=2PXp. 

Whence,  P'  =  '^  ^  ^  P ,  (  2  ) 

P+p  ^    ^ 

Again,  in  the  isosceles  triangles  ABF  and  AFM, 

Z  ABF  =  A  AFM.  (§§  193, 197.) 

Therefore,  ABF  and  AFM  are  similar.  (§  256.) 

Whence,     4^  =  ^,  or  AF"-  =  AB  X  MF. 
AB       AF 

Then  by  (1), 
Or, 


4  71" 

n              4:71                   4:71^ 

p" 

=  PX  P'. 

p' 

=  ^PX  P'. 

;0N  : 

XV.     Problem. 

Therefore,  p'  =  ^pX  P.  (3) 


376.    To  co7Jipute  aTi  approxiTnate  value  of  tt  (§  367). 

If  the  diameter  of  a  circle  is  1,  the  side  of  an  inscribed 
square  is  \  V2  (§  353)  ;  hence,  its  perimeter  is  2  V2. 

Again,  the  side  of  a  circumscribed  square  is  equal  to  the 
diameter ;  hence,  its  perimeter  is  4. 


210 


PLANE  GEOMETRY.  —  BOOK  V. 


We  then  put  in  formulae  (2)  and  (3),  Prop.  XIV., 
P  =  4,  and  j9  =  2  V2  =  2.82843. 
2PXp 


Whence, 


P'  = 


P^-P 


3.31371, 


and  y  =  V>  X  P'  =  3.06147 ; 

which  are  the  perimeters  of  the  regular  circumscribed  and 

inscribed  octagons. 

Repeating  the  operation  with  these  values,  we  put 
P  =  3.31371,  and  p  =  3.06147. 

Whence,  P'=  ^  ^  ^  -^  =  3.18260, 

P^-p 

and  y  =  V>  XP'  =  3.12145 ; 

which  are  the  perimeters  of  the  regular  circumscribed  and 

inscribed  polygons  of  sixteen  sides. 

In  this  way,  we  form  the  following  table  : 


No.  OF 

Perimeter  of 

Perimeter  of 

Sides. 

Giro.   Polygon. 

Insc.  Polygon. 

4 

4. 

2.82843 

8 

3.31371 

3.06147 

16 

3.18260 

3.12145 

32 

3.15172 

3.13655 

64 

3.14412 

3.14033 

128 

3.14222 

3.14128 

256 

3.14175 

3.14151 

512 

3.14163 

3.14157 

The  last  result  shows  that  the  circumference  of  a  circle 
whose  diameter  is  1  is  greater  than  3.14157,  and  less  than 
3.14163. 

Hence,  an  approximate  value  of  tt  is  3.1416,  correct  to  the 
fourth  decimal  place.         

Ex.  33.  The  diagonals  AG,  BD,  CE,  DF,  EA,  and  FB,  of  a 
regular  hexagon  ABCDEF,  form  a  regular  hexagon  whose  area  is 
equal  to  one-third  the  area  of  ABCJ)EF, 


MEASUREMENT   OF   THE  CIRCLE.  211 


EXERCISES. 

34.  Find  the  circumference  and  area  of  a  circle  whose  diameter 

is  5. 

35.  Find  the  radius  and  area  of  a  circle  whose  circumference  is 
75.3984. 

36.  Find  the  diameter  and  circumference  of  a  circle  whose  area 
is  201.0624. 

If  r  represents  the  radius,  a  the  apothem,  s  the  side,  and  k  the 
area,  prove  that 

37.  In  a  regular  octagon, 

a  =  ^  r  V2  +  v^,  s  =  r  V2  -  \/2,  and  A;  =  2  r2  ^/2.  (§375.) 

38.  In  a  regular  dodecagon, 

a  =  i  r  V2  +  V3,  s  =  r  V2  -  Vs,  and  fc  =  3  r^. 

39.  In  a  regular  octagon, 

r  =  a  V4  -  2  V2,  s  =  2  a  (  V2  -  1),  and  A:  =  8  a2(  V2 -1). 

40.  In  a  regular  dodecagon, 

r  =  2  a  V2-\/3,  s  =  2  a  (2  -  V3),  and  Jfc  =  12  a^  (2  -  V3). 

41.  In  a  regular  decagon,  a  =  ir  v  10  +  2  VS. 

42.  Given  one  side  of    a  regular  pentagon,   to   construct   the 
pentagon. 

43.  Given  one  side  of  a  regular  hexagon,  to  construct  the  hexagon.  * 

44.  In  a  given  square,  to  inscribe  a  regular  octagon. 

45.  In  a  given  equilateral  triangle  to  inscribe  a  regular  hexagon. 

46.  In  a  given  sector  whose  central  angle  is  a  right  angle,  to 
inscribe  a  square. 

47.  The  area  of  the  square  inscribed  in  a  sector  whose  central 
angle  is  a  right  angle  is  equal  to  one-half  the  square  of  the  radius. 

48.  The  square  inscribed  in  a  semicircle  is  equivalent  to  two-fifths 
of  the  square  inscribed  in  the  entire  circle. 

49.  If  the  diameter  of  a  circle  is  48,  what  is  the  length  of  an  arc 
of  85°? 

50.  If  the  radius  of  a  circle  is  3  Vs,  what  is  the  area  of  a  sector 
whose  central  angle  is  152°  ? 

51.  If  the  radius  of  a  circle  is  4,  what  is  the  area  of  a  segment 
whose  arc  is  120°  ? 


212  PLANE   GEOMETKY.  — BOOK  V. 

52.  Find  the  area  of  the  circle  inscribed  in  a  square  whose  area 
is  13. 

53.  Find  the  area  of  the  square  inscribed  in  a  circle  whose  area 
is  113.0976. 

54.  If  the  apothem  of  a  regular  hexagon  is  6,  what  is  the  area  of 
its  circumscribed  circle  ? 

55.  If  the  length  of  a  quadrant  is  1,  what  is  the  diameter  of  the 
circle  ? 

56.  The  length  of  the  arc  subtended  by  a  side  of  a  regular  in- 
scribed dodecagon  is  4.1888.     What  is  the  area  of  the  circle  ? 

57.  The  perimeter  of  a  regular  hexagon  circumscribed  about  a 
circle  is  12\/3.     What  is  the  circumference  of  the  circle  ? 

58.  The  area  of  a  regular  hexagon  inscribed  in  a  circle  is  24^/3. 
What  is  the  area  of  the  circle  ? 

59.  The  side  of  an  equilateral  triangle  is  6.  Find  the  areas  of 
its  inscribed  and  circumscribed  circles. 

60.  The  side  of  a  square  is  8.  Find  the  circumferences  of  its 
inscribed  and  circumscribed  circles. 

61.  Find  the  area  of  a  segment  having  for  its  chord  a  side  of  a 
regular  inscribed  hexagon,  if  the  radius  of  the  circle  is  10. 

62.  A  circular  grass-plot,  100  ft.  in  diameter,  is  surrounded  by 
a  walk  4  ft.  wide.     Find  the  area  of  the  walk. 

63.  Two  plots  of  ground,  one  a  square  and  the  other  a  circle, 
contain  each  70686  sq.  ft.  How  much  longer  is  the  perimeter  of  the 
square  than  the  circmnference  of  the  circle  ? 

64.  A  wheel  revolves  55  times  in  travelling  820.743  ft.  What  is 
its  diameter  in  inches  ? 

65.  What  is  the  number  of  degre^  in  an  arc  whose  length  is 
equal  to  that  of  the  radius  of  the  circle  ? 

66.  Find  the  side  of  a  square  equivalent  to  a  circle  whose  diam- 
eter is  3. 

67.  Find  the  radius  of  a  circle  equivalent  to  a  square  whose  side 
is  10. 

68.  If  a  circle  be  circumscribed  about  a  right  triangle,  and  on 
each  of  its  legs  as  a  diameter  a  semicircle  be  described  exterior  to  the 
triangle,  the  sum  of  the  areas  of  the  crescents  thus  formed  is  equal 
to  the  area  of  the  triangle. 

Note.    For  additional  exercises  on  Book  V.,  see  p.  228, 


APPENDIX  TO  PLANE  GEOMETRY. 


MAXIMA  AND   MINIMA    OF    PLANE    FIGURES. 


Proposition  I.     Theorem. 

377.    Of- all  triangles  formed  with  two  given  sides,  that  in 
which  these  sides  are  perpendicular  is  the  maximum. 


In  the  triangles  ABC  and  A'BC,  let  AB  =  A'B;  and  let 
AB  be  perpendicular  to  BC. 

To  prove  area  ABC  >  area  A'BG, 

Draw  A'D  perpendicular  to  BC. 

Then,  A'B  >  A'D.  (§  45.) 

That  is,  AB>A'D.  (1) 

Multiplying  both  members  of  (1)  by  ^  ^C,  we  have 
\bC  XAB>  ^BC  xA'D. 


Whence, 


area  ABC  >  area  A'BC. 


(§  313.) 


378.    Def.    Two  figures  are  said  to  be  isoperimetric  when 
they  have  equal  perimeters. 

213 


214 


PLANE   GEOMETRY.  —  APPENDIX. 


Proposition  II.     Theorem. 

379.    Of  isoperimetric  triangles  having  the  same  base,  that 
which  is  isosceles  is  the  maxirnum. 


Let  ABC  and  A'BC  be  isoperimetric  triangles  having  the 
same  base  BC;  and  let  the  triangle  ABC  be  isosceles. 
To  prove  area  ABC  >  area  A'BC. 

Produce  BA  to  D,  making  AD  =  AB,  and  draw  CD. 

Then,  Z  BCD  is  a  right  angle  ;  for  it  can  be  inscribed  in 
a  semicircle  whose  centre  is  A,  and  radius  AB.  (§  196.) 

Draw  AF  and  A^G  perpendicular  to  CD. 

Take  the  point  £J  on  CD  so  that  A'£J  =  A'C,  and  draw  BE. 

Then   since   the   triangles  ABC  and  A'BC  are   isoperi- 
metric, 

AB-^AC  =  A'B-{-  A'C  =  A'B  +  A'E. 

Whence,      A'B  +  A'E  =  AB -{- AD  =  BD. 

But,  A'B  +  A'E  >  BE.  (Ax.  6.) 

Whence,  BD  >  BE. 

Therefore,  CD  >  CE.  (§  51.) 

Now  AF  and  ^'(^  are  the  perpendiculars  from  the  vertices 
to  the  bases  of  the  isosceles  triangles  ACD  and  A'CE. 

Whence,  CF  =  ^  CD, 

and  CG  =  \  CE.  '  (§  92.) 

Therefore,  CF  >  CG.  .    (1) 

Multiplying  both  members  of  (1)  by  \  BC,  we  have 
iBCxCF>  ^BCXCG. 

Whence,  area  ABC>  area  A'BC.  (§  313.) 


MAXIMA  AND  MINIMA  OF  PLANE  FIGURES.     215 

380.  CoK.  Of  Isoperimetric  triangles,  that  which  is  equi- 
lateral is  the  maximum. 

For  if  the  maximum  triangle  is  not  isosceles  when  any- 
side  is  taken  as  the  base,  its  area  can  be  increased  by  mak- 
ing it  isosceles.  (§  379.) 

Therefore,  the  maximum  triangle  is  equilateral. 

Proposition  III.     Theorem. 

381.  Of  isoperimetric  polygons  having  the  same  number 
of  sides,  that  which  is  equilateral  is  the  maximum. 


Let  ABODE  be  the  maximum  of  polygons  having  the 
given  perimeter  and  the  given  number  of  sides. 
To  prove  that  ABODE  is  equilateral. 

If  possible,  let  the  sides  AB  and  5C  be  unequal. 

Let  AB'O  be  an  isosceles  triangle  with  the  base  AO,  hav- 
ing its  perimeter  equal  to  that  of  the  triangle  ABO. 

Then,  area  AB'0>  area  AB 0.  (§  379.) 

Adding  area  AODE  to  both  members,  we  have 
area  AB ' ODE  >  area  ABODE. 

But  by  hypothesis,  ABODE  is  the  maximum  of  polygons 
having  the  given  perimeter. 

Therefore,  area  AB'ODE  cannot  exceed  area  ABODE, 
and  hence  AB  and  BO  cannot  be  unequal. 

In  like  manner,  we  may  prove 

BO=OD  =  DE,  etc. 

•Whence,  ABODE  is  equilateral. 


216  PLANE   GEOMETRY.— APPENDIX. 

Proposition  IV.     Theorem. 

382.  Of  isoperimetric  equilateral  polygons  having  the 
same  number  of  sides,  that  which  is  equiangular  is  the 
maximum. 

O 


Let  A-F  be  the  maximum  of  equilateral  polygons  having 
the  same  perimeter  and  the  same  number  of  sides. 
To  prove  that  A-F  is  equiangular. 

If  possible,  let  Z  FAB  he  greater  than  Z  ABC. 
Produce  FA  and  CB  to  meet  at  6^.       . 
Then,  since  Z  GAB<  Z  GBA,  GA  >  GB.  (§  97.) 

Lay  ofe  GH=  GB,  and  GK  =  GA,  and  draw  HK. 
Then,  A  GRK  =  A  GAB.  (§  63.) 

Taking  away  the  triangle  GHK  from  the  entire  figure, 
there  remains  the  polygon  HKCDEF]  and   taking   away 
the  triangle  GAB,  there  remains  the  polygon  ABCDEF. 
Hence,  HKCDEF  o  AB  CDEF. 

Again,  from  the  equal  triangles  GHK  and  GAB,  we  have 
HK  =  AB.  (1) 

And  since  GA  =  GK,  and  GH  =  GB,  we  have 
GA—GH=GK—  GB,  or  AH  =  BK. 
Whence, 

FH  +  OK  =  AF  +  AH  i-  BC  -  BK 

=  AF-\-BC.  (2) 

Prom  (1)  and  (2),  HKCDEF  amd  ABCDEF  sltb  isoperi- 
metric. 


MAXIMA   AND   MINIMA  OF  PLANE   FIGURES.      217 

Then,  TIKCDEF,  being  equivalent  to  ABCDEF,  is  the 
maximum  of  polygons  having  the  given  perimeter. 

Therefore,  HKCDEF  is  equilateral.  (§  381.) 

But  this  is  impossible,  since  FH  is  greater  than  CK. 
Hence,  Z  FAB  cannot  be  greater  than  Z  ABC. 
Similarly,  Z  7^^^  cannot  be  less  than  Z.  ABC.     . 
Therefore,  Z.  FAB  =Z  ABC;  and  in  like  manner,- 

Z  ABC  =  Z  BCD  =  Z  CDF,  etc. 
Whence,  A-F  is  equiangiiliir. 

383.  CoR.    Of  isoperimefric  polygons   havivg   the   same 
number  of  sides,  that  which  is  regular  is  the'  maximicm. 

Proposition  V.     Theorem. 

384.  Of  two  isoperimefric  regular  polygons,  that  which 
has  the  greater  number  of  sides  has  the  greater  area. 


Let  ABC  be  an  equilateral  triangle,  and  N  an  isoperi- 
metric  square. 

To  prove  area  N  >  area  ABC. 

Let  D  be  any  point  in  the  side  AC  oi  the  triangle. 

Then  the  triangle  ABC  may  be  regarded  as  a  quadri- 
lateral having  the  four  sides  AB,  BC,  CD,  and  DA. 

Whence,  area  N  >  area  ABC.  (§  383.) 

In  like  manner,  we  may  prove  that  the  area  of  a  regular 
pentagon  is  greater  than  that  of  an  isoperimetric  square ;  etc. 

385.  Cor.  The  area  of  a  circle  is  greater  than  the  area 
of  any  polygon  having  an  equal  perimeter. 


218  PLANE  GEOMETRY.  —  APPENDIX. 

SYMMETRICAL  FIGURES. 
DEFINITIONS. 

386.  Two  points  are  said  to  be  syvimetrlcal  with  respect 
to  a  third,  called  the  centre  of  symmetry,  when  the  latter 
bisSc^s  the  straight  line  which  joins  them. 

Thus,  if  0  is  the  middle  point  of  the  straight  line  AB, 
the  points  A  and  B  are  symmetrical  with  a  q  jf 
respect  to  0  as  a  centre.  j \ l 

387.  The .  distance  of  a  point  from  the  centre  of  sym- 
metry is  called  the  radius  of  symmetry. 

388.  Two  points  are  said  to  be  symmetrical  with  respect 
to  a  straight  line,  called  the  axis^  of  sym- 
metry, when  the  latter  bisects  at  right 
angles   the    straight   line   which    joins 
them. 

Thus,  if  the  line  CD  bisects  AB  at     " 
right  angles,   the  points   A  and  B  are 
•symmetrical  with  respect  to  CD  as  an 
axis.  B 

389.  Two  geometrical  figures  are  said  to  be  symmetrical 
with  respect  to  a  centre,  or  with  respect  to  an  axis,  when  to 
every  point  of  one  there  corresponds  a  symmetrical  point  in 
the  other. 

In  two  such  figures,  the  corresponding  parts  are  called 
homologous. 

Thus,  if  to  every  point  of  the  triangle 
ABC  there  corresponds  a  symmetrical 
point  of  the  triangle  A'B'C,  with  respect 
to  the  centre  0,  the  triangle  A'B'C  is 
symmetrical  to  ABC  with  respect  to  the 
centre  0. 

In  this  case,  the  homologous  sides  are 
AB  and  A'B'  BC  and  B'C\  and  CA  and  C'A'. 


D 


SYMMETRICAL   FIGURES. 


219 


Again,  if  to  every  point  of  the  triangle  ABC  there  corre- 
sponds a  symmetrical  point  of  the  tri- 
angle A'B'C'f  with  respect  .to  the  axis 
DE,  the  triangle  A'B'C  is  symmetri- 
cal to  ABC  with  respect  to  the  axis 
DE. 


390.  A  figure  is  said  to  %e,  symmet- 
rical with  respect  to  a  centre  when 
every  straight  line  drawn  through  the 

centre  cuts  the  figure  in  two  points  which  are  symmetrical 
with  respect  to  that  centre. 

391.  A  figure  is  said  to  be  symmetrical  with  respect  to 
an  axis  when  it  divides  it  into  two  figures  which  are  sym- 
metrical with  respect  to  that  axis. 

Thus,  a  circumference  is  symmetrical  with  respect  to  its 
centre  as  a  centre,  or  with  respect  to  any  diameter  as  an 
axis. 

Proposition  VI.     Theorem. 

392.  Two  straight  lines  which  are  symmetrical  with  re- 
spect to  a  centre  are  equal  and  parallel. 


O" 


Let  the  straight  lines  AB  and  A'B'  be  symmetrical  with 
respect  to  the  centre  0. 

To  prove  that  AB  and  A'B'  are  equal  and  parallel. 

Draw  AA',  BB%  AB',  and  A'B. 

Then,  0  bisects  A  A'  and  BB'.  (§  386.) 

Therefore,  AB'A'B  is  a  parallelogram.  (§  111.) 

Whence,  AB  and  A'B'  are  equal  and  imrallel.  (§  104.) 


220 


PLANE   GEOMETRY.  — APPENDIX. 


Proposition  YII.     Theorem. 

393.  If  a  figure  is  symmetrical  with  respect  to  two  axes 
at  right  angles  to  each  other,  it  is  symmetrical  with  respect 
to  their  intersection  as  a  centre. 


p 

n 

A 

^ 

z__.. 

\p 

R    /\^ 

D 

H 

> 

/ 

< 

0 

E 

G 

F 

-X 


Let  the  figure  A-H  be  symmetrical  with  respect  to  the 
axes  JOT'  and  YY%  intersecting  each  other  at  right  angles 
at  0. 

To  prove  that  A-J£  is  symmetrical  with  respect  to  0  as 
a  centre. 

Let  P  be  any  point  in  the  perimeter  of  A-IT. 

Draw  FQ  and  Pi2. perpendicular  to  XX'  and  YY'. 

Produce  FQ  and  FE  to  meet  the  perimeter  of  A-H  at 
P'  and  F",  and  draw  QE,  OF',  and  OP^ 

Then,  since  A-H  is  symmetrical  with  respect  to  XX', 

FQ  =  F'Q.  (§888.) 

But  FQ  =  OE,  and  hence  OE  is  equal  and  parallel  to  F'Q. 

Therefore,  OF'QE  is  a  parallelogram.  (§  109.) 

Whence,  QE  is  equal  and  parallel  to  OF'.  (§  104.) 

In  like  manner,  we  may  prove  OF"EQ  a  parallelogram; 
and  therefore  QE  is  equal  and  parallel  to  OF". 

Hence,  since  both  OF'  and  OF"  are  equal  and  parallel  to 
QE,  F'OF"  is  a  straight  line  which  is  bisected  at  0. 

That  is,  every  straight  line  drawn  through  0  is  bisected 
at  that  point ;  whence,  A-H  is  symmetrical  with  respect  to 
0  as  a  centre.  (§  390.) 


ADDITIONAL   EXERCISES.  221 

ADDITIONAL    EXERCISES. 
BOOK  I. 

1.  The  bisectors  of  the  exterior  angles  of  a  triangle  form  a  tri- 
angle whose  angles  are  respectively  the  half-sums  of  the  angles  of 
the  given  triangle  taken  two  and  two. 

2.  If  CD  is  the  perpendicular  from  C  to  the  side  AB  of  the  tri- 
angle ABC,  and  CE  is  the  bisector  of  the  angle  C,  prove  that  ZDCE 
is  one-half  the  difference  of  the  angles  A  and  B. 

3.  The  lines  joining  the  middle  points  of  the  adjacent  sides  of  a 
quadrilateral  form  a  parallelogram  whose  perimeter  is  equal  to  the 
sum  of  the  diagonals  of  the  quadrilateral. 

4.  The  lines  joining  the  middle  points  of  the  opposite  sides  of  a 
quadrilateral  bisect  each  other. 

5.  The  lines  joining  the  middle  points  of  the  opposite  sides  of  a 
quadrilateral  bisect  the  line  joining  the  middle  points  of  the  diagonals. 

6.  The  line  joining  the  middle  points  of  the  diagonals  of  a  trape- 
zoid is  parallel  to  the  bases  and  equal  to  one-half  their  difference. 

7.  If  D  is  any  point  in  the  side  AC  of  the  triangle  ABC,  and  E, 
Fj  G,  and  //  are  the  middle  points  of  AD,  CD,  BC,  and  AB  respec- 
tively, prove  that  EFGII  is  a  parallelogram. 

8.  If  E  and  G  are  the  middle  points  of  the  sides  AB  and  CD  of 
the  quadrilateral  ABCD,  and  2^  and  II  the  middle  points  of  the 
diagonals  AC  and  BD,  prove  that  AEFH=AFGII. 

9.  If  D  and  E  are  the  middle  points  of  the  sides  BC  and  ^C  of 
the  triangle  ABC,  and  AD  be  produced  to  F  and  BE  to  G  making 
DF  =  AD  and  EG  =  BE,  prove  that  the  line  FG  passes  through  C. 

10.  If  D  is  the  middle  point  of  the  side  BC  of  the  triangle  ABC, 
prove  AD<^(iAB-{-AC). 

11.  The  sum  of  the  medians  of  a  triangle  is  less  than  the  perim- 
eter, and  greater  than  the  semi-perimeter  of  the  triangle.  (Ex.  113, 
p.  69.) 

12.  If  the  bisectors  of  the  interior  angle  at  C  and  the  exterior 
angle  aUB  of  the  triangle  ABC  meet  at  D,  prove  ZBDC  =\  ZA. 

13.  If  AD  and  BD  are  the  bisectors  of  the  exterior  angles  at  the 
extremities  of  the  hypotenuse  of  the  right  triangle  ABC,  and  DE 
and  DF  are  drawn  perpendicular,  respectively,  to  CA  and  CB  pro- 
duced, prove  that  CEDE  is  a  square. 


222  PLANE   GEOMETRY.— APPENDIX. 

14.  AD  and  BE  a,re  drawn  from  two  of  the  vertices  of  a  triangle 
ABC  to  the  opposite  sides,  making  ZBAD^ZABE  ;  if  AD  =  BE, 
prove  that  the  triangle  is  isosceles. 

•     15.   If  perpendiculars  AE,  BE,  CG,  and  DH,  be  drawn  from  the 
vertices  of  a  parallelogram  ABCD  to  any  line  in  its  plane,  prove  that 
AE-\-CG  =  BF+  DIL 

16.  If  CB  is  the  bisector  of  the  angle  C  of  the  triangle  ABC,  and 
DF  be  drawn  parallel  to  ^C  meeting  BC  at  E  and  the  bisector  of  the 
angle  exterior  to  C  at  F,  prove  that  DE  =  EF. 

17.  If  E  and  F  are  the  middle  points  of  the  sides  AB  and  AC  oi 
the  triangle  ABC,  and  AB  is  the  perpendicular  from  Aio  BC,  prove 
that  ZEDF^ZEAF. 

18.  If  the  median  drawH  from  any  vertex  of  a  triangle  is  greater 
than,  equal  to,  or  less  than  one-half  the  opposite  side,  the  angle  at 
that  vertex  is  acute,  right,  or  obtuse. 

19.  Prove  that  the  number  of  diagonals  of  a  polygon  of  n  sides  is 

n(n  — 3) 
2 

20.  The  sum  of  the  medians  of  a  triangle  is  greater  than  three- 
fourths  the  perimeter  of  the  triangle. 

21.  If  the  lower  base  AD  of  a  trapezoid  ABCD  is  double  the 
upper  base  BC,  and  the  diagonals  intersect  at  E,  prove  that 
CE^^AC  and  BE=iBD.     . 

22.  If  O  is  the  point  of  intersection  of  the  bisectors  of  the  angles 
of  the  equilateral  triangle  ABC,  and  OD  and  OE  be  drawn  respec- 
tively perpendicular  to  BC  and  parallel  to  AC,  meeting  BCatD  and 
E,  prove  that  DE  =IBC. 

23.  If  an  equiangular  triangle  be  constructed  on  each  side  of  a 
triangle,  the  lines  drawn  from  their  outer  vertices  to  the  opposite 
vertices  of  the  triangle  are  equal. 

24.  If  two  of  the  medians  of  a  triangle  are  equal,  the  triangle  is 
isosceles. 

BOOK  11. 

25.  AB  and  AC  are  the  tangents  to  a  circle  from  the  point  A, 
and  D  is  any  point  in  the  smaller  of  the  two  arcs  subtended  by  BC. 
If  a  tangent  to  the  circle  at  D  meets  AB  at  E  and  AC  at  F,  prove 
that  the  perimeter  of  the  triangle  AEF  is  constant. 

26.  The  line  joining  the  middle  points  of  the  arcs  subtended  by 
the  sides  AB  and  J. C  of  an  inscribed  triangle  ABC  cuts  AB  at  F 
and  J.C  at  (?.     Prove  that  ^F=  ^G^. 


4 


ADDITIONAL  EXERCISES.  223 

27.  If  A  BCD  is  a  circumscribed  quadrilateral,  prove  that  the 
angle  between  the  lines  joining  the  opposite  points  of  contact  is 
equal  to  i  (^ +  C). 

28.  If  the  sides  AB  and  BO  of  an  inscribed  hexagon  ABCBEF 
are  parallel  to  the  sides  DE  and  EF  respectively,  prove  that  the 
side  AF  is  parallel  to  CI). 

29.  If  AB  is  the  common  chord  of  two  intersecting  circles,  and 
AC  and  AD  are  the  diameters  drawn  from  A,  prove  that  the  line 
CD  pas%e8  through  B. 

30.  If  ^B  is  a  common  tangent  to  two  circles  which  touch  each 
other  externally  at  C,  prove  that  ACB  is  a  right  angle. 

31.  If  AB  and  AC  are  the  tangents  to  a  circle  from  the  point  vl, 
and  B  is  any  point  on  the  circumference  without  the  triangle  ABC, 
prove  that  the  sum  of  the  angles  ABD  and  ACD  is  constant. 

32.  If  A,  C,  B,  and  D  are  four  points  in  a  straight  line,  B  being 
between  C  and  D,  and  EF  is  a  common  tangent  to  the  circles  de- 
scribed upon  AB  and  CD  as  diameters,  prove  that  ZBAE  =  ZDCF. 

33.  ABCD  is  an  inscribed  quadrilateral,  AD  being  a  diameter  of 
the  circle.  If  O  is  the  centre,  and  the  sides  AD  and  BC  produced 
meet  at  E  making  CE  =  OA,  prove  that  ZAOB  =  S  Z  CED. 

34.  If  ABCD  is  an  inscribed  quadrilateral,  and  its  sides  AD  and 
BC  are  produced  to  meet  at  P,  the  tangent  at  P  to  the  circle  cir- 
cumscribed about  the  triangle  ABP  is  parallel  to  CD. 

35.  ABCD  is  a  quadrilateral  inscribed  in  a  circle.  If  the  sides 
AB  and  DC  produced  intersect  at  E,  and  the  sides  AD  and  BC  pro- 
duced at  F,  prove  that  the  bisectors  of  the  angles  E  and  F  are  per- 
pendicular to  each  other. 

36.  ABCD  is  a  quadrilateral  inscribed  in  a  circle.  Another  circle 
is  described  upon  AD  as  a  chord,  meeting  AB  and  CD  at  E  and  F. 
Prove  that  the  chords  BC  and  EF  are  parallel. 

37.  If  ABCDEFGH  is  an  inscribed  octagon,  the  sum  of  the 
angles  J.,  C,  E,  and  G  is  equal  to  six  right  angles. 

38.  If  the  number  of  sides  of  an  inscribed  polygon  is  even,  the 
sum  of  the  alternate  angles  is  equal  to  as  many  right  angles  as  the 
polygon  has  sides  less  two. 

39.  If  the  opposite  angles  of  a  quadrilateral  are  supplementary, 
the  quadrilateral  can  be  inscribed  in  a  circle. 

40.  The  perpendiculars  from  the  vertices  of  a  triangle  to  the 
opposite  sides  are  the  bisectors  of  the  angles  of  the  triangle  formed 
by  joining  the  feet  of  the  perpendiculars.     (Ex.  39. ) 


224  PLANE  GEOMETRY.— APPENDIX. 

Constructions. 

41.  Given  a  side,  an  adjacent  angle,  and  the  radius  of  the  cir- 
cumscribed'circle  of  a  triangle,  to  construct  the  triangle. 

42.  To  describe  a  circle  of  given  radius  tangent  to  a  given  circle 
and  passing  through  a  given  point. 

43.  Given  an  angle  of  a  triangle,  its  bisector,  and  the  length  of 
the  perpendicular  from  its  vertex  to  the  opposite  side,  to  construct 
the  triangle. 

44.  To  draw  between  two  given  intersecting  lines  a  straight  line 
which  shall  be  equal  to  one  given  straight  line,  and  parallelto  another. 

45.  Given  an  angle  of  a  triangle,  and  the  segments  of  the  opposite 
side  made  by  the  perpendicular  from  its  vertex,  to  construct  the 
triangle. 

46.  To  draw  a  parallel  to  the  side  BC  oi  the  triangle  ABC  meet- 
ing AB  and  AC  in  JD  and  E,  so  that  BE  may  be  equal  to  EC. 

47.  To  draw  a  parallel  to  the  side  BC  of  the  triangle  ABC  meet- 
ing AB  and  AC  in  D  and  E,  so  that  BE  may  be  equal  to  the  sum 
of  BB  and  CE. 

48.  Given  an  angle  of  a  triangle,  the  perpendicular  from  the  ver- 
tex of  another  angle  to  the  opposite  side,  and  the  radius  of  the 
circumscribed  circle,  to  construct  the  triangle. 

49.  Given  the  base  of  a  triangle,  an  adjacent  angle,  and  the  sum 
of  the  other  two  sides,  to  construct  the  triangle. 

§0.  Given  the  base  of  a  triangle,  an  adjacent  angle,  and  the 
difference  of  the  other  two  sides,  to  construct  the  triangle. 

51.  Through  a  given  point  without  a  given  circle  to  draw  a  secant 
whose  internal  and  external  segments  shall  be  equal.  (Ex.  67,  p.  103. ) 

52.  Given  the  feet  of  the  perpendiculars  from  the  vertices  of  a 
triangle  to  the  opposite  sides,  to  construct  the  triangle.     (Ex.  40.)    • 

BOOK  III. 

53.  State  and  prove  the  converse  of  Prop.  XXVI.,  III. 

54.  In  any  triangle,  the  product  of  any  two  sides  is  equal  to  the 
product  of  the  segments  of  the  third  side  formed  by  the  bisector  of 
the  exterior  angle  at  the  opposite  vertex,  minus  the  square  of  the 
bisector.     (§288.) 

55.  If  the  sides  of  a  triangle  are  AB  =  4,  AC  =  6,  and  BC  =  6, 
find  the  length  of  the  bisector  of  the  exterior  angle  at  the  vertex  A. 
(§  250.) 


ADDITIONAL  EXERCISES.  225 

56.  ABC  is  an  isosceles  triangle.  If  the  perpendicular  to  AB  at 
A  meets  the  base  BC\  produced  if  necessary,  at  E,  and  D  is  the 
middle  point  of  BE,  prove  that  AB  is  a  mean  proportional  between 
BC  and  BB.     (Ex.  88,  p.  68.) 

57.  If  D  and  E,  F  and  Gr,  and  //  and  K  are  points  on  the  sides 
AB,  BC,  and  CA,  respectively,  of  the  triangle  ABC,  so  taken  that 
AB  =  BE  =  EB,  BF  =  FG  =  GC,  and  Cll  =  JIK  =  KA,  prove 
that  the  lines  EF,  Gil,  and  KB,  when  produced,  form  a  triangle 
equal  to  ABC. 

58.  If  E  is  the  middle  point  of  one  of  the  parallel  sides  BC  of 
the  trapezoid  ABCD,  and  AE  and  BE  produced  meet  BC  and  AB 
produced  at  F  and  G,  prove  that  FG  is  parallel  to  AB. 

59.  The  perpendicular  from  the  intersection  of  the  medians  of  a 
triangle  to  any  straight  line  in  the  plane  of  the  triangle  is  one-third 
the  sum  of  the  perpendiculars  from  the  vertices  of  the  triangle  to  the 
same  line. 

60.  If  E  is  the  middle  point  of  the  median  AB  of  the  triangle 
ABC,  and  BE  meets  AC  at  F,  prove  that  AF=\AC.     (§  140.) 

61.  The  sides  AB  and  BC  of  the  triangle  ABC  are  3  and  7, 
respectively,  and  the  length  of  the  bisector  of  the  exterior  angle  B 
is  3  VT.     Find  the  side  AC. 

62.  If  three  or  more  straight  lines  divide  two  parallels  propor- 
tionally, they  pass  through  a  common  point. 

63.  The  non-parallel  sides  of  a  trapezoid  and  the  line  joining  the 
middle  points  of  the  parallel  sides,  if  produced,  meet  in  a  common 
point. 

64.  BB  is  the  perpendicular  from  the  vertex  of  the  right  angle  to 
the  hypotenuse  of  the  right  triangle  ABC.  If  E  is  any  point  in 
AB,  and  EF  be  drawn  perpendicular  to  AC,  and  FG  perpendicular 
to  AB,  prove  that  the  lines  CE  and  BG  are  parallel. 

65.  One  segment  of  a  chord  drawn  through  a  point  7  units  from 
the  centre  of  a  circle  is  4  units.  If  the  diameter  of  the  circle  is  15 
units,  what  is  the  other  segment  ? 

66.  In  a  right  triangle  ABC,  BC^  =  3  AC\  If  CB  be  drawn 
from  the  vertex  of  the  right  angle  to  the  middle  point  of  AB,  prove 
that  ZACB=  60°. 

67.  If  B  is  the  middle  point  of  the  side  BC  of  the  right  triangle 
ABC,  and  BE  he  drawn  perpendicular  to  the  hypotenuse  AB,  prove 


226  PLANE   GEOMETRY.  —  APPENDIX. 

68.  If  BE  and  CF  are  the  medians  drawn  from  the  extremities 
of  the  hypotenuse  of  the  right  triangle  ABC,  prove  that 

4  Blf+  ^CF'^^6  BCf. 

69.  If  ABC  and  ABC  are  two  angles  inscribed  in  a  semicircle, 
and  AE  and  CF  be  drawn  perpendicular  to  BD  produced,  prove 

BE'^  +  BF^  =  BE^  +  BF\ 

70.  If  perpendiculars   PF,   PD,   and   PE  be  drawn  from  any 
point  P  to  the  sides  AB,  BC,  and  CA  of  a  triangle,  prove  that 

AF^  +  BB^  +  CE-  =  AE^  +  BF^  +  CB\ 

71.  li  BC  is  the  hypotenuse  of  a  right  triangle  ABC,  prove  that 

{AB  +  BC  +  CAy  -  2  (^B  +  SC)  (BC  +  CM). 

72.  If  any  point  P  be  joined  to  the  vertices  of  the  rectangle 
ABCB,  prove  that  PA'^  +  PC'^  =  PB^  +  PB\ 

73.  If  AB  and  ^C  are  the  equal  sides  of  an  isosceles  triangle,  and 
BB  be  drawn  perpendicular  to  AC,  prove  that  2  AC  x  CB  =  BC^. 

74.  If  AB  and  BE  are  the  perpendiculars  from  the  vertices  A 
and  jB  of  the  acute-angled  triangle  ABC  to  the  opposite  sides,  prove 

AC  X  AE  +  BC  xBB  =  AB\ 

75.  The  sum  of  the  squares  of  the  diagonals  of  a  parallelogram 
is  equal  to  the  sum  of  the  squares  of  its  four  sides.     (§  279.) 


76.  To  construct  a  triangle  similar  to  a  given  triangle,  having  a 
given  perimeter. 

77.  To  construct  a  right  triangle,  having  given  its  perimeter  and 
an  acute  angle. 

78.  To  describe  a  circle  through  two  given  points,  tangent  to  a 
given  straight  line.     (§282.) 

79.  If  A  and  B  are  two  points  on  either  side  of  a  given  line  CD, 
and  AB  cuts  CB  at  F,  find  a  point  E  in  CB  such  that 

AE  :  BE  =  AF:  BF. 

BOOK  IV. 

80.  In  the  figure  on  page  176, 

(a)   Prove  that  the  lines  CF  and  BH  are  perpendicular. 
(6)    Prove  that  the  lines  AG  and  BK  are  parallel, 
(c)    Prove  that  the  sum  of  the  perpendiculars  from  n  and  L  t< 
AB  produced  is  equal  to  AB. 


ADDITIONAL  EXERCISES.  227 

(d)  Prove  that  each  of  the  triangles  AFH,  BEL,  and  CGK  is 
equivalent  to  ABC. 

(e)  Prove  that  C,  //,  and  L  are  in  the  same  straight  line. 

(/)  Prove  that  the  square  described  upon  the  sum  of  AC  and  BC 
is  equivalent  to  the  square  described  upon  the  hypotenuse,  plus  4 
times  the  area  of  ABC. 

(g)  If  EL  and  FII  be  drawn,  prove  that  the  sum  of  the  angles 
AFII,  AIIF,  BEL,  and  BLE  is  equal  to  a  right  angle. 

(h)   TTO\eth&tCF^-CE^  =  ACi^  —  BC^. 

(i)  If  FN  and  EP  are  the  perpendiculars  from  F  and  E  to  HA 
and  LB  produced,  prove  that  the  triangles  AFN  and  BEP  are  each 
equal  to  ABC. 

(j)    Prove  that  EL^  +  FH^  +  GK^  =  6  AB\ 

(fc)  Prove  that  the  lines  AL,  BU,  and  CM  meet  in  a  common 
point.     (Ex.  80,  (a).) 

(  0  Prove  that  HG,  LK,  and  MC  when  produced  meet  in  a 
common  point. 

81.  If  BE  and  CF  are  the  medians  drawn  from  the  vertices  B 
and  C  of  the  triangle  ABC,  and  intersect  at  D,  prove  that  the  tri- 
angle BCD  is  equivalent  to  the  quadrilateral  AEDF. 

82.  If  D  is  the  middle  point  of  "the  side  BC  oi  a.  triangle  ABC, 
E  the  middle  point  of  AD,  F  of  BE,  and  G  of  CF,  prove  that  ABC 
is  equivalent  to  8  EFG. 

83.  If  E  and  F  are  the  middle  points  of  the  sides  AB  and  CD 
of  a  parallelogram  ABCD,  and  AF  and  C^  be  drawn  intersect- 
ing BD  in  II  and  X,  and  iJF"  and  DE  intersecting  AC  in  K  and 
G,  prove  that  GIIKL  is  a  parallelogram  equivalent  to  ^  ABCD. 
(§  140.) 

84.  Any  quadrilateral  ABCD  is  equivalent  to  a  triangle,  two  of 
whose  sides  are  equal  to  the  diagonals  AC  and  BD  respectively, 
and  include  an  angle  equal  to  either  of  the  angles  between  AC 
and  BD. 

85.  If  similar  polygons  be  described  upon  the  sides  of  a  right 
triangle  as  homologous  sides,  the  polygon  described  upon  the  hypot- 
enuse is  equivalent  to  the  sum  of  the  polygons  described  upon  the 
legs.     (§323.) 

86.  If  through  any  point  E  in  the  diagonal  ^C  of  the  parallelo- 
gram ABCD  parallels  to  AD  and  AB  be  drawn,  meeting  AB  and 
CD  in  F  and  //,  and  BC  and  AD  in  G  and  K,  prove  that  the 
triangles  EFG  and  EHK  are  equivalent. 


228  PLANE   GEOMETRY.  — APPENDIX. 

87.  If  Ey  F,  G,  and  II  are  the  middle  points  of  the  sides  AH,  BC^ 
CD,  and  DA  of  a  square,  prove  that  the  lines  AF,  BG,  CH,  and  DE 
form  a  square  equivalent  to  \  A  BCD. 

88.  If  F  is  the  intersection  of  the  diagonals  AC  and  BD  of  a 
quadrilateral,  and  the  triangles  ABE  and  CDF  are  equivalent, 
prove  that  the  sides  AD  and  BC  are  parallel. 

89.  If  F  is  any  point  in  the  side  BC  of  the  parallelogram  ABCD, 
and  DE  be  drawn  meeting  AB  produced  at  F,  prove  that  the 
triangles  ABE  and  CFF  are  equivalent. 

90.  If  D  is  any  point  in  the  side  AB  of  a  triangle  ABC,  find  a 
point  F  in  ^C  such  that  the  triangle  ADE  is  equivalent  to  one-half 
the  triangle  ABC. 

91.  Find  the  area  of  a  trapezoid  whose  parallel  sides  are  28  and 
36,  and  non-parallel  sides  15  and  17,  respectively. 

BOOK   V. 

92.  The  area  of  the  ring  included  between  two  concentric  circles 
is  equal  to  the  area  of  the  circle  whose  diameter  is  that  chord  of  the 
outer  circle  which  is  tangent  to  the  inner. 

93.  Prove  that  an  equilateral  polygon  circumscribed  about  a 
circle  is  regular  if  the  number  of  its  sides  is  odd. 

94.  Prove  that  an  equiangular  polygon  inscribed  in  a  circle  is 
regular  if  the  number  of  its  sides  is  odd. 

95.  Prove  that  if  the  radius  of  the  circle  is  1,  the  side,  apothem, 
and  diagonal  of  a  regular  inscribed  pentagon  are 

W  (10  -  2  \/5),  i  (1  -H  VS),  and  W  (10  +  2  Vs). 

96.  The  square  of  the  side  of  a  regular  inscribed  pentagon,  minus 
the  square  of  the  side  of  a  regular  inscribed  decagon,  is  equal  to  the 
square  of  the  radius. 

97.  The  sum  of  the  perpendiculars  drawn  to  the  sides  of  a  regu- 
lar polygon  from  any  point  within  the  figure  is  equal  to  the  apothem 
multiplied  by  the  number  of  sides  of  the  polygon. 

98.  In  a  given  eq-uilateral  triangle  to  inscribe  three  equal  circles, 
tangent  to  each  other  and  to  the  sides  of  the  triangle. 

99.  In  a  given  circle  to  Inscribe  three  equal  circles,  tangent  to 
each  other  and  to  the  given  circle. 


1^ 


SOLID   GEOMETRY, 


BOOK  VI 


LINES  AND  PLANES  IN  SPACE. 
POLYEDRALS. 


DIEDRAL^.- 


394.  Def.  a  plane  is  said  to  be  determined  by  certain 
lines  or  points  when  one  plane,  and  only  one,  can  be  drawn 
through  these  lines  or  points. 


395. 

I. 

II. 

IIL 

IV. 


Proposition  I.     Theorem. 

A  plane  is  determined 

By  a  straight  line  and  a  point  without  the  line. 

By  three  points  not  in  the  same  straight  line. 

By  tido  intersecting  straight  lines. 

By  two  parallel  straight  lines. 


•  0 

/     . 

/I 

u/ 

/ 

I.   Let  C  be  a  point  without  the  straight  line  AB. 

To  prove  that  a  plane  is  determined  (§  394)  by  AB  and  C. 

If  any  plane,  as  MN,  be  drawn  through  AB,  it  may  be  re- 
volved about  AB  as  an  axis  until  it  contains  the  point  C. 

Hence,  on^  plane,  and  only  one,  can  be  drawn  through 
AB  and  C. 

229 


230  SOLID  GEOMETRY.— BOOK  VI. 


II.  Let  A,  B,  and  C  be  three  points  not  in  the  same 
straight  line. 

To  prove  that  a  plane  is  determined  by  A,  B,  and  C. 

Draw  AB. 

By  I.,  one  plane,  and  only  one,  can  be  drawn  through  the 
line  AB  and  the  point  C. 

Hence,  one  plane,  and  only  one,  can  be  drawn  through 
A,  B,  and  C. 


III.    Let  AB  and  ^C  be  two  intersecting  straight  lines. 
To  prove  that  a  plane  is  determined  by  AB  and  BC. 

By  I.,  one  plane,  and  only  one,  can  be  drawn  through  AB 
and  any  point  C  oi  BC. 

But  since  this  plane  contains  the  points  B  and  C,  it 
must  contain  the  line  BC. 

[A  plane  is  a  surface  such  that  the  straight  line  joining  any  two 
of  its  points  lies  entirely  in  the  surface.]  (§8.) 

Hence,  one  plane,  and  only  one,  can  be  drawn  through 
AB  and  BC.  .   . 


IV.   Let  AB  and  CD  be  two  parallel  lines. 

To  prove  that  a  plane  is  determined  by  AB  and  CD. 

The  parallels  AB  and  CD  lie  in  the  same  plane  (§  52). 

And  by  I.,  but  one  plane  can  be  drawn  through  AB  and 
any  point  C  of  CD. 

Hence,  one  plane,  and  only  one,  can  be  drawn  through 
AB  and  CD. 


LINES  AND   PLANES  IN  SPACE. 


231 


Proposition   II.     Theorem. 
The  intersection  of  two  vl(W£s  is  a  straight  line. 


Let  the  line  AB  be  the  intersection  of  the  planes  MN 
and  PQ. 

To  prove  AB  a  straight  line. 

Let  a  straight  line  be  drawn  between  the  points  A  and  B. 
This  line  must  lie  in  3IN,  and  also  in  PQ. 

[A  plane  is  a  surface  such  that  the  straight  line  joining  any  two 
of  its  points  lies  entirely  in  the  surface.]  (§8.) 

Then  it  must  be  the  intersection  of  MN  and  PQ. 
Whence,  AB  is  a  straight  line. 

397.  Dep.  If  a  straight  line  meets  a  plane,  the  point  of 
intersection  is  called  the  foot  of  the  line. 

A  straight  line  is  said  to  be  pei'pendlcular  to  a  plane 
when  it  is  perpendicular  to  every  straight  line  drawn  in 
the  plane  through  its  foot. 

A  straight  line  is  said  to  be  parallel  to  a  plane  when  it 
cannot  meet  the  plane  however  far  they  may  be  produced. 

Two  planes  are  said  to  be  parallel  to  each  other  when 
they  cannot  meet  however  far  they  may  be  produced. 

398.  ScH.  The  following  form  of  the  second  definition 
of  §  397  is  given  for  convenience  of  reference  : 

A  perpendicular  to  a.  'plane  is  perpendicular  to  every 
straight  line  drawn  in  the  plane  through  its  foot. 


232 


SOLID   GEOMETRY.  —  BOOK  VI. 


Proposition  III.     Theorem. 

i99.   At  a  given  point  in  a  plane,  one  perpendicular  to  tJie 
plane  can  he  drawn,  and  but  one. 


Let  P  be  the  given  point  in  the  plane  MN. 
To  prove  that  a  perpendicular  can  be  drawn  to  MN  at  P, 
and  but  one. 

At  any  point  A  of  the  straight  line  AB  draw  the  lines 
AC  and  AD  perpendicular  to  AB. 

Let  RS  be  the  plane  determined  hj  AC  and  AD. 

Let  AE  be  any  other  straight  line  drawn  through  the 
point  A  in  the  plane  RS\  and  draw  the  line  CED  inter- 
secting A  G,  AE,  and  AD  in  C,  E,  and  D. 

Produce  BA  to  B',  making  AB'  =  AB. 

Draw  BC,  BE,  BD,  B'C,  B'E,  and  B'D. 

In  the  triangles  BCD  and  B'CD,  the  side  CD  is  common. 

And  since  AC  and  AD  are  perpendicular  to  BB'  at  its 
middle  point, 

BC  =  B'C,  and  BD  =  B'D. 

[If  a  perpendicular  be  erected  af  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the 
extremities  of  the  line.]  (§  40,  I.) 

Whence,       .  A  BCD  =  A  B'CD. 

[Two  triangles  are  equal  when  the*  three  sides  of  one  are  equal 
respectively  to  the  three  sides  of  the  other.]  (§69.) 


LINES   AND  PLANES  IN  SPACE.  233 

Now  revolve  triangle  BCD  about  CD  as  an  axis  until  it 
coincides  with  triangle  B'CD. 

Then  B  will  fall  at  B' ,  and  the  line  BE  will  coincide 
with  B'E\  that  is,  BE  =  B'E. 

Hence,  since  the  points  A  and  E  are  each  equally  distant 
from  B  and  B' ^  AE  is  perpendicular  to  BB'. 

[Two  points,  each  equally  distant  from  the  extremities  of  a  straight 
line,  determine  a  perpendicular  at  its  middle  point.]  (§43.) 

But  AE  is  any  straight  line  drawn  through  A  in  BS. 
Then,  AB  is  perpendicular  to  every  straight  line  dra^n 
through  its  foot  in  the  plane  BS. 

Whence,  AB  \^  perpendicular  to  BS. 

[A  straight  line  is  said  to  be  perpendicular  to  a  plane  when  it  is 
perpendicular  to  every  straight  line  drawn  in  the  plane  through  its 
foot.]  (§397.) 

Now  apply  the  plane  B8  to  the  plane  MN  so  that  the 
point  A  shall  fall  at  P;  and  let  AB  take  the  position  FQ. 

Then,  BQ  will  be  perpendicular  to  MN. 

Hence,  a  perpendicular  can  be  drawn  to  MN  at  P. 

If  possible,  let  Bf  be  another  perpendicular  to  MN 
at  F;  and  let  the  plane  determined  by  P^  and  PT  inter- 
sect MN  in  the  line  -^/f. 

Then,  both  PQ  and  PT  are  perpendicular  to  UK. 

[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

But  in  the  plane  HKT,  only  one  perpendicular  can  be 
drawn  to  HK  at  P. 

[At  a  given  point  in  a  straight  line,  but  one  perpendicular  to  the 
line  can  be  drawn.]  (§28.) 

Hence,  but  one  perpendicular  can  be  drawn  to  MN  at  P. 

400.  Cor.  I.  A  straight  line  perpendicular  ■  to  each  of 
two  straight  lines  at  their  point  of  intersection  is  perpen- 
dicular to  their  plane. 

401.  Cor.  II.  From  a  given  point  without  a  plane,  one 
perpendicular  to  the  plane  can  he  drawn,  and  hut  one. 


234 


SOLID  GEOMETK Y.  —  BOOK  VI. 


■  The   latter  statement  is   proved 
as  follows : 

If  possible,  let  AB  and  AC  he 
two  perpendiculars  from  A  to  the 
plane  MN. 

Draw  BC'j  then  the  triangle  ABC  will  have  two  right 
angles. 

[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

But  this  is  impossible. 

Then  but  one  perpendicular  can  be  drawn  from  A  to  MN. 

402.  CoR.  III.    The  perpendicular  is   the  shortest   line 
that  can  be  drawn  from  a  point  to  a  plane. 

Let  AB  be  the  perpendicular  from  A  to  the  plane  JfiV, 
and  AC  any  other  straight  line  from  A  to  MN. 
To  prove  AB  <  AC. 

Draw  BC;  then,  since  AB  is  perpendicular  to  BCj 
AB<AC. 

[The  perpendicular  is  the  shortest  line  that  can  be  drawn  from  a 
point  to  a  straight  line.]  (§  45.) 

403.  ScH.    The  distance  of  a  point  from  a  plane  signifies 
the  length  of  "the  perpendicular  from  the  point  to  the  plane. 


Proposition  IV.     Theorem. 

404.    All  the  perpendicular's  to  a  straight  line  at  a  given 
point  lie  in  a  plane  perpoidicular  to  the  line. 


Let  AC  and  AD  be  perpendicular  to  the  line  AB  at  A. 


LINES  AND  PLANES  IN  SPACE.  235 

Then  the  plane  MN,  determined  hy  AC  and  AD,  is  per- 
pendicular to  AB. 

[A  straight  line  perpendicular  to  each  of  two  straight  lines  at  their 
point  of  intersection  is  perpendicular  to  their  plane.]  (§  400.) 

Let  AE  be  any  other  perpendicular  to  AB  at  A. 
To  prove  that  AE  lies  in  MN. 

Let  the  plane  determined,  by  AB  and  AE  intersect  MN 
in  AE']  then,  ^^  is  perpendicular  to  AE'. 

[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

But  in  the  plane  ABE,  but  one  perpendicular  can  be 
drawn  to  AB  at  A. 

[At  a  given  point  in  a  straight  line,  but  one  perpendicular  to  the 
line  can  be  drawn.]  (§28.) 

Then,  AE'  and  AE  coincide,  and  AE  lies  in  the  plane  MN 

405.  CoR.  I.  Through  a  given  point  in  a  straight  line,  a 
plane  can  he  drawn  perpendicular  to  the  line,  and  hut  one. 

406.  Cor.  II.  Through  a  given  point  without  a  straight 
line,  a  ptlane  can  he  drawn  perpendicular  to  the  line,  and  hut 
one. 

Let  C  be  the  given  point  with- 
out the  straight  line  AB. 

To  prove  that  a  plane  can  be 
drawn  through  C  perpendicular  to 
AB,  and  but  one. 

Draw  CB  perpendicular  to  AB,  and  let  BD  be  any  other 
perpendicular  to  AB  at  B. 

Then  the  plane  determined  by  BD  and  BC  will  be  a 
plane  drawn  through  C  perpendicular  to  AB. 

[A  straight  line  perpendicular  to  each  of  two  straight  lines  at  their 
point  of  intersection  is  perpendicular  to  their  plane.]  (§  400. ) 

But  only  one  perpendicular  can  be  drawn  froni  C  to  AB. 
Hence,  but  one  plane  can  be  drawn  through  C  perpen- 
dicular to  ^^. 


236 


SOLID   GEOMETRY.  —  BOOK  YI. 


Proposition  V.     Theorem. 

407.    If  oblique  lines  be  drawn  from  a  point  to  a  plane, 
I.    Two  oblique  lines  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular  from  the  point  to  the  plane  are  equal. 

II.    Of  two  oblique  lines  cutting  off  unequal  distances  from, 
the  foot  of  the  perpendicular,  the  more  remote  is  the  greater. 


I.  Let  the  oblique  lines  A  C  and  AD  meet  the  plane  MN 
at  equal  distances  from  the  foot  of  the  perpendicular  AB. 

To  prove  AC  =  AD. 

Draw  BC  and  BD. 

In  the  triangles  ABC  and  ABD,  the  side  AB  is  common. 

Also,  ZABC  =  ZABD. 

[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§398.) 

And  by  hypothesis,       BC  =  BD. 

Then,  A  ABC  =  A  ABD. 

[Two  triangles  are  equal  when  two  sides  and  the  included  angle  of 
one  are  equal  respectively  to  two  sides  and  the  included  angle  of  the 
other.]  (§63.) 

Whence,  AC  =  AD. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§66.) 

II.  Let  the  line  AU  meet  MN  at  a  greater  distance  from 
^thanJC- 

To  prove  AE  >  AC. 

Draw  BE\  on  BE  take  BF  =■-  BC,  and  draw  AF. 


LINES  AND   PLANES   IN  SPACE.  237 

Then,  AF=Aa 

[If  oblique  lines  be  drawn  from  a  point  to  a  plane,  two  oblique 
lines  cutting  off  equal  distances  from  the  foot  of  the  perpendicular 
from  the  point  to  the  plane  are  equal.]  (§  407,  I.) 

But,  AU  >  AF. 

[If  oblique  lines  be  drawn  from  a  point  to  a  straight  line,  of  two 
oblique  lines  cutting  off  unequal  distances  from  the  foot  of  the  per- 
pendicular, the  more  remote  is  the  greater.]  (§  48,  II.) 

Whence,  AF  >  AC. 

Proposition  VI.     Theorem. 

408.  (Converse  of  Prop.  V.,  I.)  Tivo  equal  oblique  lines 
froin  a  point  to  a  plane  cut  off  equal  distances  from  the  foot 
of  the  perpendicular  from  the  point  to  the  plane. 


Let  AC  and  AD  be  equal  oblique  lines,  and  AB  the  per- 
pendicular, from  A  to  the  plane  MN]  and  draw  BC  and  BD. 
To  prove  BC  =  BD. 

In  the  triangles  ABC  and  ABD,  AB  is  common. 
And  by  hypothesis,       AC  =  AD. 
Also,  ABC  and  ABD  are  right  angles. 
[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

Whence,  A  ABC  =  A  ABD. 

[Two  right  triangles- are  equal  when  the  hypotenuse  and  a  leg  of 
one  are  equal  respectively  to  the  hypotenuse  and  a  leg  of  the  other.] 

(§88.) 
Therefore,  BC  =  BD.    ■  ' 


238  SOLID   GEOMETRY.— BOOK  VI. 

409.  Cor.  (Converse  of  Prop.  V.,  II.)  If  two  unequal 
oblique  lines  be  drawn  from  a  point  to  a  plane,  the  greater 
cuts  off  the  greater  distance  from  the  foot  of  the  perpen- 
dicular from  the  point  to  the  plane. 

(The  proof  is  left  to  the  student.) 

Proposition  VII.     Theorem. 

410.  If  through  the  foot  of  a  perpendicular  to  a  plane  a 
line  be  drawn  at  right  angles  to  any  line  in  the  plane,  the 
line  drawn  from  its  intersection  with  this  line  to  any  point 
in  the  perpendicular  will  be  perpendicular  to  the  line  in  the 
plane. 

B 


Let  AB  be  perpendicular  to  the  plane  MN. 
Draw  AE  perpendicular  to  any  line  CI>  in  MN,  and  join 
E  to  any  point  B  in  AB. 

To  prove  BE  perpendicular  to  CD. 

On  CD  take  EC  =  ED-,  and  draw  AC,  AD,  BC,  and  BD. 
Then,  AC  =  AD. 

[If  a  perpendicular  be  erected  at  the  middle  point  of  a  straight 
line,  any  point  in  the  perpendicular  is  equally  distant  from  the  ex- 
tremities of  the  line.]  (§  40,  I.) 

Therefore,  BC  =  BD. 

[If  oblique  lines  be  drawn  from  a  point  to  a  plane,  two  oblique 
lines  cutting  off  equal  distances  from  the  foot  of  the  perpendicular 
from  the  point  to  the  plane  are  equal.]  (§  407,  I.) 

Whence,  BE  is  perpendicular  to  CD. 

[Two  points,  each  equally  distant  from  the  extremities  of  a  straight 
line,  determine  a  perpendicular  at  its  middle  point.]  (§  43.) 


J 


LINES  AND  PLANES  IN  SPACE.  239 

Proposition  VIII.     Theorem. 
/;41 1  -    Two  perpendiculars  to  the  same  plane  are  parallel, 

A\<:        o 


Let  the  lines  AB  and  CD  be  perpendicular  to  the  plane 

MN. 

To  prove  AB  and  CD  parallel. 

♦Let  A  be  any  point  of  AB,  and  draw  AD  and  BD. 
Also,  draw  DF  in  th4  plane  JfiV  perpendicular  to  BD. 
.    Then  Ci)  is  perpendicular  to, DF. 

[A  perpendicular  to  a  plane  is  perpendicular  to  evetf^traight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

Also,  AD  is  perpendicular  to  DF. 

[If  through  the  foot  of  a  perpendicular  to  a  plane  a  line  be  drawn 
at  right  angles  to  any  line  in  the  plane,  the  line  drawn  from  its  inter- 
section with  this  line  to  any  point  in  the  perpendicular  will  be  per- 
pendicular to  the  line  in  the  plane.]  (§  410.) 

Therefore,  CD,  AD,  and  BD,  being  perpendicular  to  DF 
at  D,  lie  in  the  same  plane. 

[All  the  perpendiculars  to  a  straight  line  at  a  given  point  lie  in  a 
plane  perpendicular  to  the  line.]  (§  404.) 

Hence,  AB  and  CD  lie  in  the  same  plane. 

[A  plane  is  a  surface  such  that  the  straight  line  joining  any  two 
of  its  points  lies  entirely  in  the  surface.]  (§8.) 

Again,  AB  and  CD  are  perpendicular  to  BD. 

[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

Whence,  AB  and  CD  are  parallel. 

[Two  perpendiculars  to  the  same  straight  line  are  parallel.]  (§54.) 


240 


SOLID   GEOMETRY.  —  BOOK  VI. 


412.  Cor.  I.    If  one  of  two  parallels  is  jperpendicular  to  a 
plane,  the  other  is  also  perpendicular  to  the  plane. 

Let  the  lines  AB  and  CD  be  par- 
allel, and  let  AB  be  perpendicular  to 
the  plane  MN.  m 

To  prove  CD  perpendicular  to  MK. 

A  perpendicular  from  C  to  JfiV 
will  be  parallel  to  AB. 

[Two  perpendiculars  to  the  same  plane  are  parallel.]  (§  411.) 

But  through  C,  only  one  parallel  can  be  drawn  to  AB. 

[But  one  straight  line  can  be  drawn  through  a  given  point  parallel 
to  a  given  straight  line.]  (§  53.) 

Whence,  CD  is  perpendicular  to  MJV. 

413.  Cor.  II.    If  each  of  two  straight  lines  is  parallel  to 
a  third,  they  areijarallel  to  each  other. 

Let  the  lines  AB  and  CD  be  paral- 
lel to  EF. 

To  prove  AB  and  CD  parallel. 

Draw  the  plane  MN  perpendicular 
\.oEF. 

Then  each  of  the  lines  AB  and  CD 
'is  perpendicular  to  MN. 

[If  one  of  two  parallels  is  perpendicular  to  a  plane 
also  perpendicular  to  the  plane.] 

Whence,  AB  and  CD  are  parallel. 


the  other  is 

(§412.) 


[Two  perpendiculars  to  the  same  plane  are  parallel.] 


(§  411.) 


EXERCISES. 

1.  What  is  the  locus  (§  141)  of  the  perpendiculars  to  a  given 
straight  line  AB  at  the  point  A  ? 

2.  What  is  the  locus  of  points  equally  distant  from  the  circumfer- 
ence of  a  given  circle  ? 

3.  If  a  plane  bisects  a  straight  line  at  right  angles,  any  point  in 
the  plane  is  equally  distant  from  the  extremities  of  the  line. 


A 


LINES   AND   PLANES   IN   SPACE. 


241 


Proposition  IX.     Theorem. 

414.    A  straight  line  parallel  to  a  line  in  a  plane  is  paral- 
lel to  the  plane. 


Let  AB  he  parallel  to  the  line  CD  in  the  plane  MN. 
To  prove  AB  parallel  to  MN. 

The  parallels  AB  and  CD  lie  in  a  plane,  which  intersects 
MN  in  the  line  CD, 

Hence,  if  AB  meets  MN,  it  must  be  at  some  point  of  CD. 

But  AB,  being  parallel  to  CD,  cannot  meet  it. 

Then  AB  and  MN  cannot  meet,  and  are  parallel  (§  397). 


Proposition   X.     Theorem. 

418)  If  a  straight  line  is  parallel  to  a  plane,  the  intersec- 
tion of  the  plane  with  any  plane  drawn  through  the  line  is 
parallel  to  the  line. 


A 

M 

B 

/ 

/ 

/  G 

/ 

is 

Let  the  line^^  be  parallel  to  the  plane  MN;  and  let  CD_ 
be  the  intersection  of  MN  with  a  plane  drawn  through  AB. 
To  prove  AB  and  CD  parallel. 

The  lines  AB  and  CD  lie  in  the  same  plane. 
And  since  AB  cannot  meet  the  plane  MN  however  far 
they  may  be  produced,  it  cannot  meet  CD. 
Therefore,  AB  and  CD  are  parallel  (§  52). 


242 


SOLID   GEOMETRY.  —  BOOK   VI. 


A 

M 

B 

' 

/ 

/  C. 

I 
/ 

)/ 

JV 

416.    Cor.    If  a  line  and  a  jjlane  are  parallel,  a  parallel 
to  the  line  through  any  point  of  the 
plane  lies  in  the  plane. 

Let  the  line  AB  hQ  parallel  to  the 
plane  MN]  and  through  any  point  C 
of  MN  draw  CD  parallel  to  AB. 

To  prove  that  CD  lies  in  MN. 

The  plane  determined  by  AB  and  C  intersects  MN  in  a 
parallel  to  AB. 

[If  a  straight  line  is  parallel  to  a  plane,  the  intersection  of  the 
plane  with  any  plane  drawn  through  the  line  is  parallel  to  the  line.] 

(§415.) 

But  through  C,  only  one  parallel  can  be  drawn  to  AB. 
[But  one  straight  line  can  be  drawn  through  a  given  point  parallel 
to  a  given  straight  line.]  (§53.) 

Whence,  CD  lies  in  MN. 


Proposition  XI.     Theorem. 

417.    If  two  parallel  planes  are  cut  by  a  third  plane,  the 
intersections  are  parallel. 


Let  the  parallel  planes  MN  and  P^  be  cut  by  the  plane 
AD  in  the  lines  AB  and  CD,  respectively. 
To  prove  AB  and  CD  parallel. 

The  lines  AB  and  CD  lie  in  the  same  plane. 
And  since  the  planes  MN  and  PQ  cannot  meet  however 
far  they  may  be  produced,  AB  and  CD  cannot  meet. 
Therefore,  AB  and  CD  are  parallel  (§  52). 


LINES   AND  PLANES   IN   SPACE.  243 

418.    Coil.    Parallel  lines  included  between  parallel  planes 
are  equal. 

Let  AC  and  BD  be  parallel  lines  included  between  the 
parallel  planes  MN  and  PQ. 

To  prove  AC  =  BD. 

Let  the  plane  determined  hy  AC  and  BD  intersect  MN 
and  PQ  in  the  lines  AB  and  CD. 
Then,  AB  and  CD  are  parallel. 

[if  two  parallel  planes  are  cut  by  a  third  plane,  the  Intersections 
are  parallel.]  (§417.) 

Therefore,  AC  =  BD. 

[Parallel  lines  included  between  parallel  lines  are  equal.]  (§  105.) 

Proposition  XII.     Theorem. 

Two  planes  perpendicular  to  the  same  straight  line 
are  parallel. 


N 


Let  the  planes  MN  and  PQ  hQ  perpendicular  to  the 
line  AB. 

To  prove  MN  and  PQ  parallel. 

If  MN  and  PQ  are  not  parallel,  they  will  meet  if  suffi- 
ciently produced  ;  let  C  be  a  point  in  their  intersection. 

There  Avill  then  be  two  planes  drawn  through  C  perpen- 
dicular to  AB,  which  is  impossible. 

[Through  a  given  point  without  a  straight  line,  but  one  plane  can 
be  drawn  perpendicular  to  the  line.]  (§  406.) 

Whence,  MN  and  PQ  cannot  meet,  and  are  parallel. 


244  SOLID    GEOMETRY.— BOOK  VI. 


Proposition   XIII.     Theorem. 

420.    If  each  of  two   intersecting  lines  is  iiarallel   to  a 
plane,  their  plane  is  parallel  to  the  given  plane. 


Let  AB  and  AC  he  parallel  to  the  plane  PQ. 
To  prove  their  plane  MN  parallel  to  FQ. 

Draw  AD  perpendicular  to  P^. 

Through  D  draw  DE  and  DP  parallel  to  AB  and  AG. 

Then,  DE  and  DF  lie  in  the  plane  FQ. 

[If  a  line  aiid  a  plane  are  parallel,  a  parallel  to  the  line  through 
any  point  of  the  plane  lies  in  the  plane.]  (§  416.) 

Whence,  AD  is  perpendicular  to  DE  and  DF. 
[A  perpendicular  to  a  plane  is  pei-pendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§398.) 

Therefore,  AD  is  perpendicular  to  AB  and  AC. 
[A  straight  line  perpendicular  to  one  of  two  parallels  is  perpen- 
dicular to  the  other.]  (§56.) 

Hence,  AD  is  perpendicular  to  MN. 

[A  straight  line  perpendicular  to  each  of  two  straight   lines  at 
their  point  of  intersection  is  perpendicular  to  their  plane;]     (§  400.) 

Then  3IN  and  P(>  are  parallel. 

[Two  planes  perpendicular  to  the  same  straight  line  are  parallel.] 

(§419.) 
EXERCISES. 

4.  A  line  parallel  to  each  of  two  intersecting  planes  is  parallel  to 
their  intersection. 

5.  A  straight  line  and  a  plane  perpendicular  to  the  same  straight 
line  are  parallel. 


^ 


LINES  AND  PLANES  IN  SPACE.  245 

Proposition  XIV.     Theorem. 

421.    A  straight  line  i)ei"pend'wular  to  one  of  two  parallel 
planes  is  perpendicular  to  the  other  also. 


M 

/     2^:^>^<C^^  / 

p 

N 

/  » 

Q 

Let  MN  and  PQ  he  parallel  planes ;  and  let  the  line  AD 
be  perpendicular  to  FQ. 

To  prove  AD  perpendicular  to  MN. 

Pass  any  two  planes  through  AD,  intersecting  MJ^  in  AD 
and  AC,  and  P^  in  DE  and  DF. 

Then  AB  is  parallel  to  DB,  and  AC  to  DF. 

[If  two  parallel  planes  are  cut  by  a  third  plane,  the  intersections 
are  parallel.]  (§417.) 

But  AD  is  perpendicular  to  DF  and  DF. 

[A  perpendicular  to  a  plane  is  perpendicular  to  every  straight  line 
drawn  in  the  plane  through  its  foot.]  (§  398.) 

Whence,  AD  is  perpendicular  to  AB  and  AC. 

[A  straight  line  perpendicular  to  one  of  two  parallels  is  perpen- 
dicular to  the  other.]  (§56.) 

Therefore,  AD  is  perpendicular  to  MN. 

[A  straight  line  perpendicular  to  each  of  two  straight  lines  at 
their  point  of  intersection  is  perpendicular  to  their  plane.]     (§  400.) 

422.    Cor.  I.   Two  parallel  planes  are  everywhere,  eqnalhj 
distant  (§  403). 

For  all  common  perpendiculars  to  the  planes  are  parallel. 
[Two  perpendiculars  to  the  same  plane  are  parallel.]  (§  411.) 

Therefore  they  are  all  equal. 
[Parallel  lines  included  between  parallel  planes  are  equal.]  (§  418.) 


246 


SOLID   GEOMETRY.  —  BOOK  VI. 


M, 


423.    Cor.  II.    Through   a  given  point  a  plane  can  he 
drawn  parallel  to  a  given  plane,  and  hut  one. 

Let  A  be  the  given  point,  and  TQ 
the  given  plane. 

To  prove  that  a  plane  can  be  drawn 
through  A  parallel  to  TQ,  and   but  P, 

one. 


^N 


Draw  ^^  perpendicular  to  P^. 

Through  A  pass  the  plane  MN  perpendicular  to  AB. 
Then  MN  will  be  parallel  to  P^. 

[Two  planes  perpendicular  to  the  same  straight  line  are  parallel.] 

(§419.) 
If  another  plane  could  be  drawn  through  A  parallel  to 
P^,  it  would  be  perpendicular  to  AB. 

[A  straight  line  perpendicular  to  one  of  two  parallel  planes  is  per- 
pendicular to  the  other  also.]  (§421.) 

It  would  then  coincide  with  MN. 

[Through  a  given  point  in  a  straight  line,  but  one  plane  can  be 
drawn  perpendicular  to  the  line.]  (§  405.) 

Then  but  one  plane  can  be  drawn  through  A  parallel  to  PQ. 

pROPOSiTioif  XV.     Theorem. 

424.  If  two  angles  not  in  the  same  2>l<^'fi&  have  their  sides 
parallel  and  extending  in  the  same  direction,  they  are  equal, 
and  their  planes  are  parallel. 


Let  MN  and  P^  be  the  planes  of  the  angles  BAC  and 
B'A!^C'  \  and  let  AB  and  ^C  be  parallel  respectively  to 
AlB'  and  A'C ,  and  extend  in  the  same  direction. 


J 


LINES   AND  PLANES    IN  SPACE.  247 

I.  To  prove  /.  BAC  =  Z  B'A'C\ 

Lay  off  AB  =  A'B',  and  AC  =  A'C';  and  draw  ^^', 
BB',  CC\BC,  SLudB'C'. 

Then  since  AB  is  equal  and  parallel  to  A'B',  the  figure 
ABB' A'  is  a  parallelogram. 

[If  two  sides  of  a  quadrilateral  are  equal  and  parallel,  the  figure 
is  a  parallelogram.]  (§  109.) 

Whence,  AA'  is  equal  and  parallel  to  BB'. 
[The  opposite  sides  of  a  parallelogram  are  equal.]  (§  104.) 

In  like  manner,  AA'  is  equal  and  parallel  to  CC". 
Therefore,  BB'  is  equal  and  parallel  to  CC. 

[If  each  of  two  straight  lines  is  parallel  to  a  third,  they  are  parallel 
to  each  other.]  (§413.) 

Whence,  BB'C'C  is  a  parallelogram,  and  BC  =  B'C. 
Therefore,  A  ABC  =  A  A'B'C 

[Two  triangles  are  equal  when  the  three  sides  of  one  are  equal 
respectively  to  the  three  sides  of  the  other.]  (§  69.) 

Whence,  Z.  BAG  =  Z.  B'A'C. 

[In  equal  figures,  the  homologous  parts  are  equal.]  (§  66.) 

II.  To  prove  MN  parallel  to  PQ. 

The  lines  AB  and  AC  are  each  parallel  to  the  plane  PQ. 

[A  straight  line  parallel  to  a  line  in  a  plane  is  parallel  to  the 
plane.]  (§414.) 

Therefore,  MN  is  parallel  to  P^.   . 

[If  each  of  two  intersecting  lines  is  parallel  to  a  plane,  their  plane 
is  parallel  to  the  given  plane.]  (§  420.) 

EXERCISES. 

6.  What  is  the  locus  of  points  equally  distant  from  a  given  plane  ? 

7.  If  two  planes  are  parallel,  a  line  parallel  to  one  of  them 
through  any  point  of  the  other  lies  in  the  other. 

8.  If  two  planes  are  parallel  to  a  third  plane,  they  are  parallel  to 
each  other. 

9.  If  a  line  is  parallel  to  a  plane,  it  is  everywhere  equally  distant 
from  the  plane. 


248  SOLID   GEOMETRY.— BOOK  VI. 


Proposition  XVI.     Theorem. 

425.    If  two  straight  lines  are  cut  by  three  parallel  planes, 
the  corresponding  segments  are  proportional. 


Let  the  parallel  planes  MN,  PQ,  and  RS  intersect  the 

lines  AC  and  A'C  in  the  points  A,  B,  C,  and  A%  B',  C, 

respectively. 

^  AB      A'B' 

To  prove  ^  =  ^7^.' 

Draw  AC. 

Through  AC  and  AC^  pass  a  plane,  intersecting  BQ  and 
BS  in  the  lines  BD  and  CC. 

Then  BB  is  parallel  to  CC\ 

[If  two  parallel  planes  are  cut  by  a  third  plane,  the  intersections 
are  parallel.]  (§417.) 

•  Therefore,  |f  =  ^.  (1) 

[A  parallel  to  one  side  of  a  triangle  divides  the  other  two  sides 
proportionally.]  (§  245.) 

In  like  manner,  -^^  =  ^^ .  (2) 

From  (1)  and  (2),        |f,  =  ff. 


Ex.  10.   Through  a  given  point  a  plane  can  be  drawn  parallel  to 
any  two  straight  lines  in  space.     (§  414.) 


i 


DIEDRALS. 


249 


DIEDRALS. 
Definitions. 

426.  If  two  planes  meet  in  a  straight  line,  the  figure 
formed  is  called  a  diedral  angle,  or  simply  a 

diedral. 

The  line  of  intersection  of  the  planes  is 
called  the  edge  of  the  diedral,  and  the  planes 
are  called  its  faces. 

Thus,  in  the  diedral  formed  by  the  planes 
BD  and  BF,  BE  is  the  edge,  and  BD  and 
BF  are  the  faces. 

427.  A  diedral  may  be  designated  by  two  letters  on  its 
edge ;  or,  if  several  diedrals  have  a  common  edge,  by  four 
letters,  one  in  each  face  and  two  on  the  edge,  the  letters  on 
the  edge  being  named  between  the  other  two. 

Thus,  the  above  diedral  may  be  designated  BE,  or  ABEC. 

428.  The  plane  angle  of  a  diedral  is  the  angle  formed 
by  two  straight  lines  drawn  one  in   each 

face,  perpendicular  to  the  edge  at  the  same 
point. 

Thus,  if  the  lines  AB  and  J  C  be  drawn 
in  the  faces  DE  and  I^F,  respectively,  per- 
pendicular to  DG  at  A,  BAC  is  the  plane 
angle  of  the  diedral  DG. 

429.  Let  the  lines  A'B'  and  A'C  be  drawn  in  the  faces 
DE  and  DF,  respectively,  perpendicular  to  DG  at  A'. 

Then,  A'B'  is  parallel  to  AB,  and  A'C  to  AC.          (§  54.) 
Whence,  Z  B'A'C  =  Z  BAC.  (§  424.) 

That  is,  the  plane  angle  of  a  diedral  is  of  the  same  mag- 
nitude at  whatever  point  of  the  edge  it  may  be  drawn. 

430.  Two  diedrals  are  equal  when  their  faces  may  be 
made  to  coincide. 


250 


SOLID  GEOMETRY.  —  BOOK  VI. 


431.  It  is  evident  tliat  two  diedrals  are  equal  when  their 
pla7ie  angles  are  equal. 

^32.  Conversely,  the  plane  angles  of  equal  diedrals  are 
equal. 

433.  A  plane  perpendicular  to  the  edge  of  a  diedral  inter- 
sects the  faces  in  lines  perpendicular  to  the  edge  (§  398). 

Hence,  a  plane  perpendicular  to  the  edge  of  a  diedral 
intersects  the  faces  in  lines  which  form  the  plane  angle  of 
the  diedral,  . 


434.  Two  diedrals  are  said  to  be  adjacent 
when  they  have  the  same  edge,  and  a  com- 
mon face  between  them;  as  ABEC  and 
CBED. 

Two  diedrals  are  said  to  be  vertical  when 
the  faces  of  one  are  the  extensions  of  the  faces  of  the  other. 

435.  Through  a  given  straight  line  in  a  plane,  a  plane 
may  be  drawn  meeting  the  given  plane  in  such  a  way  as  to 
make  the  adjacent  diedrals  equal.     (Compare  §  27.) 

Each  of  the  equal  diedrals  is  called  a  right  diedral,  and 
the  planes  are  said  to  be  perpendicular  to  each  other. 

Thus,  if  the  plane  PQ  he  drawn 
meeting  the  plane  3IN  in  such  a  way 
as  to  make  tlie  adjacent  diedrals 
PBQM  and  PRQN  equal,  each  of 
these  is  a  right  diedral,  and  MN  and 
PQ  are  perpendicular  to  each  other. 

436.  Througli  a  given  line  in  a  plane  but  one  .plane  can  be 
drawn  perpendicular  to  the  given  plane.     (Compare  §  2'^.) 

437.  The  projection  of  a  point  on  a  plane  is  the  foot  of 
the  perpendicular  drawn  from  the  point  to  the  plane. 

The  projection  of  a  line  on  a  plane  is  tlie  locus  (§  141)  of 
the  projections  of  its  points. 


A 


BIEDRALS.  251 

Proposition  XVII.     Theorem. 
438.    The  plane  angle  of  a  right  dledral  is  a  right  angle. 


3r. 


JV 

Let  the  planes  PQ  and  MN  be  perpendicular  to  each 
other,  and  intersect  in  the  line  QR. 

Let  ABC  and  ABC  be  the  plane  angles  of  the  diedrals 
FRQN  and  PRQM. 

To  prove  ABC  a  right  angle. 

Since  PQ  is  perpendicular  to  MN,  we  have 

diedral  PRQN  =  diedral  PRQM.  (§  435.) 

Whence,                  Z  ABC  =  Z  ABC.  (§  432.) 

Therefore,  JJ5C  is  a  right  angle.  (§  27.) 

439.  Cor.  (Converse  of  Prop.  XVII.)  If  the  plane  angle 
of  a  diedral  is  a  right  angle,  the  faces  of  the  diedral  are  per- 
pendicular to  each  other. 

Let  the  planes  PQ  and  MN  intersect  in  the  line  QR. 

Let  ABC  and  ABC  be  the  plane  angles  of  the  diedrals 
PRQN  and  PRQM,  and  let  ABC  he  ^  right  angle. 

To  prove  PQ  perpendicular  to  MN. 

Since  ABC  is  a  right  angle,  we  have 

ZABC=ZABC.  (§27.) 

Whence,     diedral  PRQN  =  diedral  PRQM.  (§  431.) 

Therefore,  PQ  is  perpendicular  to  MN.  (§  435.) 


Ex.  H.   Through  any  given  straight  line  a  plane  can  be  drawn 
parallel  to  any  other  straight  line.     (§  414.) 


252 


SOLID   GEOMETRY.  —  BOOK   VI. 


Proposition  XVIII.     Theorem. 

If  two  planes  are  perpendicular  to  each  other,  a 
straight  line  drawn  in  one  of  them  perpendicular  to  their 
intersection  is  perpendicular  to  the  other. 


'  Let  the  plane  PQhe  perpendicular  to  MN. 

Let  QE  be  their  intersection,  and  draw  AB  in  the  plane 
PQ  perpendicular  to  QB. 

To  prove  AB  perpendicular  to  MN. 

Draw  BC  in  the  plane  MN  perpendicular  to  QB. 
Then  ABC  is  the  plane  angle  of  the  diedral  FBQN. 

(§428.) 

Whence,  ABC  is  a  right  angle.  (§  438.) 

Therefore  AB,  being  perpendicular  to  BC  and  BQ  at  ^, 

is  perpendicular  to  the  plane  MN.  (§  400.) 

441.  Cor.  I.  If  two  planes  are  perpendicular,  a  perpen- 
dicular to  one  of  them  at  any  point  of  their  intersection  lies 
in  the  other. 

Let  the  plane  PQ  hQ  perpendicular  to  MN;  at  any  point 
B  in  their  intersection  QB,  draw  AB  perpendicular  to  MN. 

To  prove  that  AB  lies  in  PQ. 

A  line  drawn  in  P^  perpendicular  to  QB  at  B  will  be 
perpendicular  to  MN.  (§  440.) 

But  at  the  point  B,  but  one  perpendicular  can  be  drawn 
toMN.  (§399.) 

Hence,  AB  lies  in  PQ. 


DIEDRALS. 


253 


442.  Cor.  II.  If  two  planes  are  perpendicular,  a  perpen- 
dicular to  one  from  any  point  of  the  other  lies  in  the  other. 

Let  the  plane  PQ  he  perpendicular  to  MN-,  and  through 
any  point  Aoi  PQ  draw  AB  perpendicular  to  MN. 

To  prove  that  AB  lies  in  PQ. 

A  line  drawn  in  PQ  through  the  point  A,  perpendicular  to 
the  intersection  QR,  will  be  perpendicular  to  MN.    (§  440.) 

But  from  the  point  A,  but  one  perpendicular  can  be  drawn 
to  MN.  (§  401.) 

Hence,  AB  lies  in  PQ. 


Proposition  XIX.     Theorem. 

443.    If  a  straight  line  is  perpendicular  to  a  plane,  every 
plane  drawn  through  the  line  is  perpendicular  to  the  plane^ 


M 

A 

/ 

/ 

p 

/       c/ 

/ 

B      y 

Q  N 

Let  the  line  ^^  be  perpendicular  to  the  plane  MN\  and 
let  PQ  be  any  plane  drawn  through  AB. 
To  prove  PQ  perpendicular  to  MN. 

Let  QP  be  the  intersection  of  P^  and  MN,  and  draw  BC 
in  the  plane  MN  perpendicular  to  QR. 

Now  AB  is  perpendicular  to  BQ.  (§  398.) 

Then  ABC  is  the  plane  angle  of  the  diedral  PRQN. 

(§  428.) 
But  ^^C  is  a  right  angle.  (§  398.) 

Hence,  P^  is  perpendicular  to  MN.  (§  439.) 

444.    Cor.    A  plane  perpendicular  to  the  edge  of  a  diedral 
is  perpendicular  to  its  faces. 


254 


SOLID   GEOMETRY.  —  BOOK  VI. 


Proposition  XX.     Theorem.  ^ 

445.   A  plane  perpendicular  to  each  of  two  intersecting 
planes  is  perpendicular  to  their  intersection. 


Let  the  planes  FQ  and  ^>S^  be  perpendicular  to  MN. 
To  prove  their  intersection  AB  perpendicular  to  MN. 

Let  a  perpendicular  be  drawn  to  MN  at  B. 

This  perpendicular  will  lie  in  both'P^  and  US.      (§  441.) 

It  must  therefore  be  their  line  of  intersection. 

Hence,  AB  is  perpendicular  to  MJ^. 


Proposition  XXI.     Theorem. 

446.    Every  point  in  the  bisecting  plane  of  a  diedral  is 
equally  distant  from  the  faces  of  the  diedral. 


From  any  point  P  in  the  bisecting  plane  BE  of  the  die- 
dral ABDC,  draw  PM  and  PuST  perpendicular  to  AI>  and  CD. 
To  prove  PM  =  PN. 

Let  the  plane  determined  by  PM  and  PN  intersect  the 
planes  AD,  BE,  and  CD  in  EM,  FP,  and  EN. 


DIEDRALS.  255 

The  plane  PMFN  is   perpendicular  to  the  planes  AD 
and  CD.  (§  443.) 

Then  the  plane  PMFN  is  perpendicular  to  BD.    (§  445.) 
Therefore,  FFM  and  FFN  are  the  plane  angles  of  the 
diedrals  ABDE  and  CEDE.  (§  433.) 

Whence,  ,  ~~Z.  PFM  =  Z  PFN.  (§  432.) 

Now  in  the  right  triangles  PFMimii  PFN,  PF  is  common. 
Also,  ZPFM  =  Z  PFN 

Therefore,  A  PFM  =  A  PFN,  (§  70.) 

Whence,  PM  =  PN  (§  66.) 

Proposition  XXII.     Theorem. 

447.    If  two  planes   intersect,  the  vertical  diedrals  are 
equal. 


Let  the  planes  MN  and  PQ  intersect  in  the  line  BS. 
To  prove     diedral  PESN  =  diedral  MESQ. 

Let  AEC  and  A' EC  be  the  plane  angles  of  the  diedrals 
PESN  and  MESQ. 

Then,  Z  ABC  =  Z  ABC.  (§  39.) 

Whence,      diedral  PESN  =  diedral  MESQ.  (§  431.) 

EXERCISES. 

12.  If  two  parallel  planes  are  cut  by  a  third  plane,  the  alternate- 
interior  diedrals  are  equal. 

13.  If  a  straight  line  is  parallel  to  a  plane,  any  plane  perpendicu- 
lar to  the  line  is  perpendicular  to  the  plane. 

14.  If  a  plane  be  drawn  through  a  diagonal  of  a  parallelogram,  the 
perpendiculars  to  it  from  the  extremities  of  the  other  diagonal  are 
equal. 


256 


SOLID   GEOMETRY.  —  BOOK   YI. 


Proposition  XXIII.     Theorem. 

'448.  Through  a  given  straight  line  without  a  plane,  a 
plane  can  he  drawn  perpendicular  to  the  giveti  plane,  and 
but  one. 


Let  AB  be  the  given  line  witlioiit  the  plane  MN. 
To  prove  that  a  plane  can  be  drawn  through  AB  perpen- 
dicular to  MJV,  and  but  one. 

Draw  AC  perpendicular  to  MN,  and  let  AD  be  the  plane 
determined  hj  AB  and  AC. 

Then,  AD  is  perpendicular  to  MJST.  (§  443.) 

If  more  than  one  plane  could  be  drawn  through  AB  per- 
pendicular to  MN,  their  common  intersection,  AB,  would  be 
perpendicular  to  MN.  (§  445.) 

Hence,  but  one  plane  can  be  drawn  through  AB  perpen- 
dicular to  MN,  unless  AB  is  perpendicular  to  MN 

Note.  If  the  line  AB  is  perpendicular  to  MN,  an  indefinitely 
great  number  of  planes  can  be  drawn  through  AB  perpendicular  to 
MN  (§  443). 

449.  Cor.  The  projection  of  a  straight  line  on  a  plane  is 
a  straight  line. 

Let  CD  be  the  projection  of  the  straight  line  AB  on  the 
plane  MN. 

To  prove  CD  a  straight  line. 

Let  a  plane  be  drawn  through  AB  perpendicular  to  MN. 

The  perpendiculars -to  MN  from  all  points  of  AB  will  lie 

in  this  plane.  (§  442.) 

Therefore,  CD  is  a  straight  line.  (§  396.) 


DIEDRALS.  267 


Proposition  XXIV.     Theorem 

450.  The  angle  between  a  straight  line  and  its  projection 
on  a  plane  is  the  least  a.ngle  which  it  makes  with  any  line 
drawn  in  the  plane  through  its  foot. 


^ 


Let  BC  be  the  projection  of  the  line  AB  on  the  plane  MN. 
Let  BD  be  any  other  line  drawn  through  B  in  MN. 
To  prove  /.ABC  <Z  ABD. 

Lay  ofe  BD  =  BC,  and  draw  AC  and  AD. 
Then  in  the  triangles  ABC  and  ABD,  AB  is  common. 
Also,  AC<AD.  (§402.) 

Whence,  ZABC<Z  ABD.  (§  90.) 

451.   Sen.   ABC  is  called  the  angle  between  AB  and  MN. 

EXERCISES. 

15.  If  two  parallels  meet  a  plane,  they  make  equal  angles  with  it. 

16.  If  a  straight  line  intersects  two  parallel  planes,  it  makes  equal 
angles  with  them. 

17.  The  angle  between  perpendiculars  to  the  faces  of  a  diedral 
from  any  point  within  the  angle  is  the  supplement  of  its  plane  angle. 

18.  If  BC  is  the  projection  of  the  line  AB  upon  the  plane  MN, 
and  BD  and  BE  be  drawn  in  the  plane  making  Z  CBD  =  Z  CBEj 
prove  that  Z  ABD  =  Z  ABE. 

19.  If  each  of  two  intersecting  planes  be  cut  by  two  parallel 
planes,  not  parallel  to  their  intersection,  their  intersections  with  the 
parallel  planes  include  equal  angles. 

20.  The  line  AB  is  perpendicular  to  the  plane  MN  at  B.  A  line 
is  drawn  from  B  meeting  the  line  CD  of  the  plane  MN  at  E.  If 
AE  is  perpendicular  tP  C2>,  prove  that  BE  is  perpendicular  to  CD. 


268  SOLID   GEOMETRY.  —  BOOK  VI. 

>_EQEYEDRALS. 
Definitions. 

452.  If  three  or  more  planes  meet  in  a  common  point, 
the  figure  formed  is  called  a  polyedral 

angle  J  or  simply  a  polyedral.  o 

The  common  point  is  called  the  vertex 
of  the  polyedral,  and  the  intersections  of  / 

the  planes  the  edges.  / 

The  portions  of  the  planes  included'  AjL""' 
between  the  edges  are  called  the  faces  /  \J 
of  the  polyedral,  and  the  angles  formed  r 

by  the  edges  are  called  the  face  angles. 

Thus,  in  the  polyedral  0-ABCD,  0  is  the  vertex;  OA, 
OB,  etc.,  are  the  edges;  the  planes  AOB,  BOC,  etc.,  are  the 
faces;  and  the  angles  AOB,  BOC,  etc.,  are  the  face  angles. 

453.  A  polyedral  must  have  at  least  three  faces. 
A  polyedral  of  three  faces  is  called  a  triedral. 

454.  To  show  more  distinctly  the  relative  positions  of 
the  edges  of  a  polyedral,  it  is  customary  to  represent  them 
as  intersected  by  a  plane,  as  shown  in  the  figure  of  §  452. 

The  plane  ABCD  is  called  the  base  of  the  polyedral. 

455.  The  polyedral  is  not  regarded  as  limited  by  the 
base;  thus,  the  face  AOB  is  understood  to  mean,  not  the 
triangle  AOB,  but  the  indefinite  plane  included  between 
the  edges  OA  and  OB  produced  indefinitely. 

456.  A  polyedral  is  called  convex  when  its  base  is  a  con- 
vex polygon  (§  120). 

457.  Two  polyedrals  are  called  vertical  when  the  edges  of 
one  are  the  prolongations  of  the  edges  of  the  other. 


458.    Two  polyedrals  are  equal  when  they  can  be  applied 
to  each  other  so  that  their  f£K?es  shall  coincide. 


I 


POLYEDRALS.  259 

Two  polyedrals  are  equal  when  the  face  angles  and 
diedrals  of  one  are  equal  respectively  to  the  homologous 
face  angles   and   diedrals   of 

the  other,  if  the  equal  parts  ^  a 

are    arranged    in    the    same  /  \  /  \ 

order.  /  \  /   I     \ 

Thus,  if  the  face  angles  A^-4---^c  a'<~/--^(/ 
AOB,   BOC,    and    COA    are  iT  b^ 

equal  respectively  to  the  face 

angles  A'O'B',  B'O'C,  and  CO' A',  and  the  diedrals  OA,  OB, 
and  OC  to  the  diedrals  a  A',  a  B' ,  and  O'C,  the  triedrals 
0-ABC  and  O'-AB'C  are  equal;  for  they  can  evidently 
be  applied  to  each  other  so  that  their  faces  shall  coincide. 

460.  Two  polyedrals  are  said  to  be  symmetrical  when  the 
face  angles  and  diedrals  of  one  are  equal  respectively  to 
the  homologous  face  angles 

and  diedrals  of  the  other,  if  >y  a 

the  equal  parts  are  arranged  //  \  /  \\ 

in,  the  reverse  order.  /  \  /     \     \ 

Thus,  if  the  face  angles  a4--4--;^o  c'Cc^^Yx"*' 
AOB,  BOC,   and   COA  are        -Y"^^^  ^b' 

equal    respectively    to    the 

face  angles  A'OB',  B'OC,  and  COA,  and  the  diedrals 
OA,  OB,  and  OC  to  the  diedrals  0' A' ,  O'B',  and  O'C",  the 
triedrals  0-ABC  and  O^-AB'C  are  symmetrical. 

461.  It  is  evident  that,  in  general,  two  symmetrical  poly- 
edrals cannot  be  placed  so  that  their  faces  shall  coincide. 

EXERCISES. 

21.  The  three  planes  bisecting  the  diedrals  of  a  triedral  meet  in  a 
common  straight  line. 

22.  D  is  any  point  in  the  perpendicular  AF  from  A  to  the  side 
BC  of  the  triangle  ABC.  If  DE  be  drawn  perpendicular  to  the 
plane  of  ABC,  and  GH  be  drawn  through  E  parallel  to  BC,  prove 
that  AE  is  perpendicular  to  GIL     (§  398.) 


260  SOLID   GEOMETRY.— BOOK  VI. 


Proposition  XXV.     Theorem. 

462.  The  sum  of  any  two  face  angles  of  a  triedral  is 
greater  than  the  third. 

Note.  The  theorem  requires  proof  only  in  the  case  where  the 
third  angle  is  greater  than  either  of  the  others. 


In  the  triedral  0-ABC,  let  the  face  angle  AOC  be  greater 
than  either  AOB  ov  BOC. 

To  prove        A  AOB -{- Z  BOC>  /.  AOC. 

In  the  face  AOC,  draw  the  line  OD  equal  to  OB,  making 
Z  AOD  =  Z  AOB ;  and  through  B  and  D  pass  a  plane 
cutting  the  faces  of  the  triedral  in  AB,  BC,  and  CA. 

Then  in  the  triangles  AOB  and  AOD,  OA  is  common. 

And  by  construction,    OB  =  OD, 
and  ZAOB  =  ZAOD. 

Therefore,  A  AOB  =  A  A  OD.  (§  63.) 

AVhence,  AB  =  AD.  (§  66.) 

Now,  AB-\-BC>  AD-{-DC.  (Ax.  6.) 

Or,  since  AB  =  AD,     BC  >  DC. 

Then  in  the  triangles  BOC  and  COD,  OC  is  common. 

Also,  OB  =  OD,  and  BC>  CD. 

Whence,  Z  BOC  >  Z  COD.  (§  90.) 

Adding  Z  AOB  to  the  first  member  of  this  inequality, 
and  its  equal  Z  A  OD  to  the  second  member,  we  have 
Z  AOB  +  Z  BOC  >  Z  AOD  +  Z  COD. 

Whence,         Z  AOB -\- Z  BOC  >  Z  AOC. 


POLYEDRALS.  261 


Proposition  XXVI.    Theorem. 

463.    The  sum  of  the  face  angles  of  any  convex  polyedral 
is  less  than  four  right  angles. 


Let  0- ABODE  be  a  convex  polyedral. 

To  prove  ZAOB-{-Z  BOG  +  etc.  <  four  right  angles. 

Let  ABODE  be  the  base  of  the  polyedral. 

Let  0'  be  any  point  within  the  polygon  ABODE,  and 
draw  a  A,  OB,  aO,  OD,  and  a  E. 

Then,    Z  OAE  +  Z  OAB  >  Z  O'AE  +  Z  OAB.    (§  462.) 

In  like  manner, 

Z  OB  A  +  Z  05C  >  Z  0'^^  +  Z  aBO\  etc. 

Adding  these  inequalities,  we  have  the  sum  of  the  angles 
at  the  bases  of  the  triangles  whose  common  vertex  is  0 
greater  than  the  sum  of  the  angles  at  the  bases  of  the 
triangles  whose  common  vertex  is  O . 

But  the  sum  of  all  the  angles  of  the  triangles  whose  com- 
mon vertex  is  0  is  equal  to  the  sum  of  all  the  angles  of  the 
triangles  whose  common  vertex  is  Of.  (§  82.) 

Hence,  the  sum  of  the  angles  at  0  is  less  than  the  sum 
of  the  angles  at  0'. 

Therefore,  the  sum  of  the  angles  at  0  is  less  than  four 
right  angles.  (§  37.) 

Ex.  23.  Between  two  straight  lines  not  in  the  same  plane  a  com- 
mon perpendicular  can  be  drawn.     (Ex.  11.) 


262 


SOLID   GEOMETKY.  —  BOOK  VI. 


PRorosiTiON   XXVII.     Theorem. 

464.    If  two  triedrals  have  the  face  angles  of  one  equal 
respectively  to  the  face  angles  of  the  other, 

I.    They  art  equal  if  the  equal  parts  occur  in  the  same 
order. 

II.    They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 


I.    In  the  triedrals  0-ABC  and  O'-A'B'C,  let 
Z.  AOB  =  Z  A'O'B',  A  BOC  =  Z  B'aC, 
and  Z  COA=Z  CO' A'. 
To  prove  triedral  0-ABC  =  triedral  O'-A'B'C 

Lay  off  the  six  equal  distances  OA,  OB,  OC,  aA',  O'B', 
and  aC;  and  draw^^,  BC,  CA,  A'B',  B'C\  and  C"^'. 

Then,  A  OAB  =  A  O'A'B'.  (§  63.) 

Whence,  AB  =  A'B'.  (§  m.) 

Similarly,      BC  =  B'C,  and  CA  =  CA'. 

Therefore,  A  ^^C  =  A  A'B'C.  (§  69.) 

Draw  OD  and  O'l/  perpendicular  to  ABC  and  A'B'C, 
respectively;  also,  draw  AD  and  A'jy. 

The  equal  oblique  lines  OA,  OB,  and  OC  meet  the  plane 
ABC  at  equal  distances  from  D.  (§  408.) 

Hence,  D  is  the  centre  of  the  circumscribed  circle  of 
the  triangle  ABC;  and  similarly,  jy  is  the  centre  of  the  cir- 
cumscribed circle  of  A'B'C. 

Now  apply  0'- A'B'C  to  0-ABC,  so  that  the  points  A',  B', 
and  C  shall  fall  at  A,  B,  and  C,  and  the  point  J)'  at  D. 


POLYEDRALS.  263 

Then  tlie  perpendicular  O'D'  will  fall  upon  OD.    (§  399.) 

But  the  right  triangles  OAD  and  OAH/  are  equal. 

(§  88.) 

Whence,  ajy  =  OD,  and  the  point  (7  will  fall  at  0. 

Therefore,  the  triedrals  0-ABC  and  (y-A'B'C  coincide 
throughout,  and  are  equal. 

II.  In  the  triedrals  0-ABC  and  0"-A"B"C",  let  the 
angles  AOB,  BOC,  and  COA  he  equal  respectively  to 
A^'a'B",  B"a'C",  and  Ca'A!'. 

To  prove  0-ABC  symmetrical  to  a'-A'B"C". 

Construct  O-A'B'C  symmetrical  to  a'-A'B"C'\  having 
the  angles  AOB',  B'OC,  and  COA!  ,equal  respectively 
to  A!'a'B",  B"a'C",  and  CO' A'. 

Then  the  triedrals  0-ABC  and  a-AB'C  have  tlie 
angles  AOB,  BOC,  and  COA  equal  respectively  to  A!O^B\ 
B'aC,  and  CaA'. 

Hence,  triedral  0-ABC  =  triedral  O'-A'B'C',   (§  464, 1.) 

Therefore,  0-ABC  is  symmetrical  to  0"-A"B"C". 

465.  CoK.  If  two  triedrals  have  the  face  angles  of  one 
equal  respectively  to  the  face  angles  of  the  other,  their  homol- 
ogous diedrals  are  equal. 

EXERCISES. 

24.  Two  triedrals  are  equal  when  two  face  angles  and  the  included 
diedral  of  one  are  equal  respectively  to  two  face  angles  and  the  in- 
cluded diedral  of  the  other,  and  similarly  arranged. 

25.  Two  triedrals  are  equal  when  a  face  angle  and  the  adjacent 
diedrals  of  one  are  equal  respectively  to  a  face  angle  and  the  adja- 
cent diedrals  of  the  other,  and  similarly  arranged. 

26.  A  is  any  point  in  the  face  EG  of  the  diedral  BEFG.  li  AG 
be  drawn  perpendicular  to  the  edge  EF^  and  AB  perpendicular  to 
the  face  BF^  prove  that  the  plane  determined  by  J.C  and  ^C  is 
perpendicular  to  EF, 

27.  From  any  point  E  within  the  diedral  CABD,  EF  a.nd  EG  are 
drawn  perpendicular  to  the  faces  ABC  and  ABD,  and  Gil  perpen- 
dicular to  the  face  ABC  at  II.     Prove  FII  perpendicular  to  AB. 


BOOK  YIL 

POLYEDRONS. 


DEFINITIONS. 

466.  A  polyedron  is  a  solid  bounded  by  planes. 

The  bounding  planes  are  called  the  faces  of  the  polye- 
dron ;  their  intersections  are  called  the  edges,  and  the  inter- 
sections of  the  edges  the  vertices. 

A  diagonal  is  a  straight  line  joining  any  two  vertices  not 
in  the  same  face. 

467.  The  least  number  of  planes  which  can  form  a  poly- 
edral  is  three  (§  453) ;  hence,  the  least  number  of  planes 
which  can  bound  a  polyedron  is  four. 

A  polyedron  of  four  faces  is  called  a  tetraedron\  of  six 
faces,  a  hexaedron ;  of  eight  faces,  an  octaedron ;  of  twelve 
faces,  a  dodecaedron ;  of  twenty  faces,  an  icosaedron. 

468.  A  polyedron  is  called  convex  when  the  section  made 
by  any  plane  is  a  convex  polygon  (§  120). 

All  polyedrons  considered  hereafter  will  be  understood  to 
be  convex. 

469.  The  volume  of  a  solid  is  its  ratio  to  another  solid, 
called  the  unit  of  volume,  adopted  arbitrarily  as  the  unit  of 
measure  (§  179). 

470.  Two  solids  are  said  to  be  equivalent  when  their 
volumes  are  equal. 

264 


PRISMS  AND  PARALLELOPIPEDS. 


265 


PRISMS  AND   PARALLELOPIPEDS. 

471.  A  prism  is  a  polyedron,  two  of  whose  faces  are 
equal  polygons  lying  in  parallel  planes, 
having   their   homologous    sides    parallel, 
the  other  faces  being  parallelograms. 

The  equal  and  parallel  faces  are  called 
the  bases  of  the  prism,  and  the  remaining 
faces  the  lateral  faces  ;  the  intersections  of 
the  lateral  faces  are  called  tlie  lateral  edges, 
and  the  sum  of  the  areas  of  the  lateral  faces  the  lateral  area. 

The  altitude  is  the  perpendicular  distance  between  the 
planes  of  the  bases. 

472.  The  following  is  given  for  convenience  of  reference: 
The  bases  of  a  prism  are  equal. 

473.  It  follows  from  the  definition  of  §  471  that 
The  lateral  edges  of  a  2^ris7n  are  equal  and  parallel. 

474.  A  prism  is   called  triangular,  quadrangular,  etc., 
according  as  its  base  is  a  triangle,  quadrilateral,  etc. 

475.  A  right  prism  is  a  prism  whose  lat- 
eral edges  are  perpendicular  to  its  bases. 

An  oblique  prism  is  a  prism  whose  lateral 
edges  are  not  perpendicular  to  its  bases. 


476.  A   regular  prism   is   a   right    prism 
whose  base  is  a  regular  polygon. 

477.  A  truncated  prism  is  that  portion  of  a 
prism  included  between  the  base,  and  a  plane, 
not  parallel  to  the  base,  cutting  all  the  lateral 


478.    A  right  section  of  a  prism  is  the  section 
made  by  a  plane  perpendicular  to  the  lateral  edges. 


2m 


SOLID   GEOMETKY.  —  BOOK  VII. 


479.  A  2^ci'^(^ll^lopi2Jed  is  a  prism  whose 
bases  are  parallelograms ;  that  is,  all  the 
faces  are  parallelograms. 

480.  A  right  parallelopiped  is  a  par- 
allelopiped  whose  lateral  edges  are  per- 
pendicular to  its  bases. 

481.  A  rectangular  parallelopiped  is  a 
right  parallelopiped  whose  bases  are  rect- 
angles ;  that  is,  all  the  faces  are  rectangles. 

The  dimensions  are  the  three  edges  which 
meet  at  any  vertex.   . 

482.  A  cube  is  a  rectangular  parallelopiped  whose  six 
faces  are  all  squares. 

Proposition  I.     Theorem. 

483.  The  sections  of  a,  prism  made  by  two  i^arallel  planes 
which  cut  all  the  lateral  edges,  are  equal  polygons. 


Let  the  parallel  planes  CF  and  C'F'  cut  all  the  lateral 
edges  of  the  prism  AB. 

To  prove  that  the  sections  CDEFG  2indi  C'D'F'F'G'  are 
equal. 

We  have  CD  parallel  to  C'ly,  BE  to  lyE',  etc.     (§  417.) 
Whence,  CD  =  CD',  DE  =  D^E',  etc.  (§  105.) 


A 


PRISMS   AND  PARALLELOPIPEDS.  267 

Then  the  polygons  CDEFG  and  C'lyE'F'G'  are  mutu- 
ally equilateral. 

Again,  Z  CDE  =  Z  C'lyE', 

Z  DEF  =  Z  I/E'F',  etc.  (§  424.) 

Then  the  polygons  CDEFG  and  C'D'E'F'G'  are  mutu- 
ally equiangular. 

Therefore,  CDEFG  and  C'D^E'F'G'  are  equal.     (§  124.) 

484.  Coit.    The  section  of  a  prism  made  hy  a  plane  par- 
allel to  the  base  is  equal  to  the  base. 

Proposition  II.     Theorem. 

485.  The  lateral  area  of  a  prism  is  equal  to  the  perimeter 
of  a  right  section  multip)lied  by  a  lateral  edge. 


Let  DEFGH  be  a  right  section  of  the  prism  AC. 
To  prove 

lat.  area  AC  =  {DE  +  EF  +  etc.)  x  AA'. 
We  have  DE  perpendicular  to  AA'.  (§  398.) 

Whence,  area  AA'B'B  =  DE  X  AA\  (§  310.) 

Similarly,        area  BB' CC  =  EF  X  BB' 

=  EF  XAA'-,  etc.        (§  473.) 
Adding  these  equations,  we  have 

lat.  area  AC  =  DE  X  A  A'  +  EF  X  AA'  +  etc. 
=  (D^  +  ^^i'^  +  etc.)  X  ^^'. 

486.    CoR.    The  lateral  area  of  a  right  prism  is  equal  to 
the  p)erimeter  of  the  base  multiplied  by  the  altitude. 


268 


SOLID   GEOMETRY.  —  BOOK   Yil. 


Propositiox  III.     Theorem. 

487.  Two  prisms  are  equal  when  the  faces  including  a 
triedral  of  one  are  equal  respectively  to  the  faces  including 
a  triedral  of  the  other,  and  similarly  placed. 


In  the  prisms  AH  and  A'H',  let  the  faces  ABODE,  AG, 
and  AL  be  equal  respectively  to  the  faces  AE'C'IfE' ,  A'G', 
and  AL' ;  the  equal  parts  being  similarly  placed. 

To  prove  the  prisms  equal. 

The  angles  EAB,  EAF,  and  FAB  are  equal  respectively 
to  the  angles  E'AB',  E'A'F',  and  F'A'B\ 

Then,  triedral  A-BEF  =  triedral  A'-B'E'F\     (§  464,  I.) 

Then  the  prism  A'H'  may  be  applied  to  AH  so  that  the 
vertices  A\  B\  C[,  ly,  E' ,  G' ,  F' ,  and  L'  shall  fall  at  A,  B, 
C,  D,  E,  G,  F,  and  L,  respectively. 

Now  since  the  lateral  edges  of  the  prisms  are  parallel, 
the  edge  C'H'  will  fall  upon  CH,  and  D'JC  upon  DIC 

And  since  the  points  G',  F',  and  U  fall  at  G,  'F,  and  L, 
the  planes  of  the  upper  bases  will  coincide.  (§  395,  II.) 

Therefore,  the  points  H'  and  7i '  fall-  at.  H  and  K. 

Hence,  the  prisms  coincide  throughout,  and  are  equal. 

488.  ScH.  The  above  demonstration  applies  without 
change  to  the  case  of  two  truncated  prisms. 


Cor.  Two  right  prisms  are  equal  ivhen  they  have 
equal  bases  and  equal  altitudes  ;  for  by  inverting  one  of  the 
prisms  if  necessary,  the  equal  faces  will  be  similarly  placed. 


PlllSMS  AND   PAKALLELOPIPEDS.  269 


Proposition  IV.     Theorem. 

490.  An  oblique  prism  is  equivalent  to  a  right  prism, 
having  for  its  base  a  right  sectio7i  of  the  oblique  prism,  a7id 
for  its  altitude  a  lateral  edge  of  the  oblique  prism. 


Let  FGHKL  be  a  right  section  of  the  oblique  prism  Ajy. 
Produce  AA'  to  F',  making  FF'  =  AA'. 
At  i^^'pass  the  plane  F'K'  parallel  to  FGHKL,  meeting 
the  edges  BB',  CC,  etc.,  produced  at  G',  H',  etc. 
To  prove  AI/  equivalent  to  the  ri^ht  prism  FK'. 

In  the  truncated  prisms  yi/f  and  A'K',  the  faces  FGHKL 
and  F'G'H'K'L'  are  equal.  (§  472.) 

Therefore,  AK'  may  be  applied  to  AK  so  that  the  ver- 
tices F',  G',  etc.,  shall  fall  at  F,  G,  etc.,  respectively. 

Then  the  edges  A'F',  B'G',  etc.,  will  coincide  in  direction 
with  AF,  BG,  etc.  (§  399.) 

But,  since  FF'  =  AA',  we  have  AF  =  A'F'. 

In  like  manner,  BG  =  B'G',  CH  =  C'H',  etc. 

Hence,  the  vertices  A',  B',  etc.,  will  fall  at  A,  B,  etc. 

Then,  A'K'  and  AK  coincide  throughout,  and  are  equal. 

Now  taking  from  the  entire  solid  AK'  the  truncated 
prism  A'K',  there  remains  the  prism  AD^. 

And  taking  its  equal  AK,  there  remains  the  prism  FK'. 

Hence,  Ajy  and  FK'  are  equivalent. 


270 


SOLID   GEOMETRY.  —  BOOK  VII. 


Proposition  V.     Theorem. 

491.    The  opposite  faces  of  a  parallelopiped  are  equal  and 
parallel. 


Let  AC  and  A'C  be  the  bases  of  the  parallelopiped  AC. 
To  prove  the  faces  AB'  and  DC  equal  and  parallel. 

AB  is  equal  and  parallel  to  DC,  and  A  A'  to  DD'.  (§  104.) 

Hence,  Z  A'AB  =  Z.  D'DC, 

and  the  faces  AB'  and  DC  are  parallel.  (§  424.) 

Therefore,  the  faces  AB'  and  DC  are  equal.  (§  112.) 

In  like  manner,  we  may  prove  AD'  and  BC  equal  and 
parallel. 

492.  Cor.  Either  face  of  a  parallelopiped  may  he  taken 
as  the  base.  * 

Proposition  VI.     Theorem. 

493.  The  plane  passed  through  two  diagonally  opposite 
edges  of  a  parallelopiped  divides  it  into  two  equivalent  trian- 
gular prisms. 

:o' ^0' 


Let  ^C"  be  a  parallelopiped. 


{ 


PRISMS  AND  PARALLELOPIPEDS.  271 

Through  the  edges  AA'  and  CC"  pass  a  plane  dividing 
AC  into  two  triangular  prisms,  ABC-A'  and  ACD-A'. 
To  prove  ABC-A'  -o  A  CD-A'. 

Let  EFGH  be  a  right  section  of  the  parallelepiped,  cut- 
ting the  plane  AA'C'C  in  EG. 

Now  the  planes  AB'  and  DC  are  parallel.  (§  491.) 

Whence,  EF  is  parallel  to  GH.  (§  417.) 

In  like  manner,  EH  is  parallel  to  FG. 

Therefore,  EFGH  is  a  parallelogram. 

Whence,  A  EFG  =  A  EGH.  (§  106.) 

Now,  ABC-A'  is  equivalent  to  a  right  prism  whose  base 
is  EFG,  and  altitude  AA';  and  ACD-A'  is  equivalent  to  a 
right  prism  whose  base  is  EGH,  and  altitude  AA\    (§  490.) 

But  these  two  right  prisms  are  equal.  (§  489.) 

Therefore,  ABC-A'  =o  ACD-A\ 

Proposition  VII.     Theorem. 
494.    The  diagonals  of  a  parallelopiped  bisect  each  other. 


Let  AC  and  A'C  be  diagonals  of  the  parallelopiped  AC\ 
To  prove  that  AC  and  A'C  bisect  each  other. 

Drawee  and  ^'C. 

Then  AA'  is  equal  and  parallel  to  CC.  (§  473.) 

Whence,  the  figure  AA'CC  is  a  parallelogram.      (§  109.) 
Therefore,  AC  and  A'C  bisect  each  other  at  0.     (§  110.) 
In  like  manner,  we  may  prove  that  any  two  of  the  four 
diagonals  AC,  A'C,  BD',  and  B'D  bisect  each  other  at  0. 
Note.     The  point  O  is  called  the  centre  of  the  parallelopiped. 


272 


SOLID   GEOMETKY.— BOOK  VII. 


Proposition  VIII.     Theorem. 

495.    Two  rectangular  parallelopipeds  having  equal  bases 
are  to  each  other  as  their  altitudes. 

Note.     The  phrase  "rectangular  parallelopiped "  in  the  above 
statement  signifies  the  volume  of  tlie  rectangular  parallelopiped. 

Case  I.    When  the  altitudes  are  commensurable. 


/  A 

-A 

D' 

Q 

A           y 

/       Jr 

A-- 

-/-. 

/]-- 

/ 1- — 

/y 

/''     j 

/  J 

/ 

V 

/ 

/ 

Let  P  and  Q  be  two  rectangular  parallelopipeds,  having 
equal  bases,  and  commensurable  altitudes  AA!  and  BB', 

n.  P  AA' 

To  prove  —  = . 

^  Q        BB' 

Let  ^C  be  a  common  measure  of  A  A'  and  BB' ,  and  let  it 
be  contained  4  times  in  AA' ,  and  3  times  in  BB'. 

Then,  ^  =  ^.  (1) 

BB'      ^  ^    ^ 

At  the  several  points  of  division  of  AA'  and  BB'  pass 
planes  perpendicular  to  these  lines. 

Then  the  parallelopiped  P  will  be  divided  into  4  parts, 
and  the  parallelopiped  Q  into  3  parts,  all  of  which  parts 
will  be  equal.  (§  489.) 

Therefore,  —  =  -. 

Q       3 

From  (1)  and  (2),  we  have 

P  ^AA' 

Q       BB'' 


PRISMS   AND   PAllALLELOPIPEDS.  273 

Case  II.    When  the  altitudes  are  incommensurable. 


AL 


B^ 


A-- 

// 

/        \ 

/ 

1 

1 

A" 
/ 

/ 

Let  P  and  Q  be  two  rectangular  parallelepipeds,  having 
equal  bases,  and  incommensurable  altitudes  AA!  and  BB'. 

P  ^  AA' 
Q      BB'' 


To  prove 


Let  AA'  be  divided  into  any  number  of  equal  parts,  and 
let  one  of  these  parts  be  applied  to  BB'  as  a  measure. 

Since  AA'  and  BB'  are  incommensurable,  a  certain  num- 
ber of  the  parts  will  extend  from  B  to  C,  leaving  a  remainder 
CB'  less  than  one  of  the  parts. 

Pass  the  plane  CD  perpendicular  to  BB'^  and  let  Q'  denote 
the  rectangular  parallelopiped  BD. 

Then  since  A  A'  and  BC  are  commensurable, 

^  =  ^-  .(§495,  Case  I.) 

Now  let  the  number  of  subdivisions  of  AA'  be  indefinitely 
increased. 

Then  the  length  of  each  part  will  be  indefinitely  dimin- 
ished, and  the  remainder  CB'  will  approach  the  limit  0. 

P  P 

Then,  -—  will  approach  the  limit  — , 

q  (? 

and  — —  will  approach  the  limit ^. 

BC  BB 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 

P  ^  A  A' 

Q      BB'' 


Whence, 


274 


SOLID  GEOMETRY.  —  BOOK  VII. 


496.  ScH.    The  theorem  may  also  be  expressed : 

Two  rectangular  parallelopijoeds  having  two  dimensions  in 
common,  are  to  each  other  as  their  third  dimensions. 

Proposition  IX.     Theorem. 

497.  Two  rectangular  parallelopipeds  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 


P 

Q 

T» 

/       1 

c 

A 

// 

/V- 

/ 

/ 
/ 

—  - 

C 

1 

) — 

c 

/  ' 

/ 

Let  P  and  Q  be  two  rectangular  parallelopipeds,  having 
the  dimensions  a,  h,  c,  and  a',  h',  c,  respectively. 
P  ^  aXh 
Q 


To  prove 


Whence, 


a'  X  h' 

Let  i^  be  a  rectangular  parallelopiped  having  the  dimen- 
sions a',  h,  and  c. 

Then  F  and  R  have  the  dimensions  b  and  c  in  common. 

Whence,    •  C  =  ^-  (§496.) 

R       a 

And  R  and  Q  have  the  dimensions  a'  and  c  in  common. 

R^b_ 

Q   y 

Multiplying  these  equations,  we  have 
P  ^  aXb 
Q       a'xb'' 

498.    ScH.    The  theorem  may  also  be  expressed  : 

Two  rectangular  parallelopipeds  having  one  dimension  in 

comm,on,  are  to  each  other  as  the  products  of  their  other  two 

dimensions. 


PlillSMS   AND   PAKALLELOPIPEDS. 


275 


Proposition  X.     Theorem. 

499.    Any   two    rectangular  pdrallelopipeds   are   to   each 
other  as  the  products  of  their  three  dimensions. 


P 

R 

r\ 

/ 

V 

/  \                   /  \ 

/  \ 

/ 

/         / 

/\ 

/ 

/ 

/ 

C 

/            1 

/               1 

/ 

i 

y— 
/ 
/ 

c 

1  — 

/h 

/ 

c'      / 

A 

Let  P  and  Q  be  two  rectangular  parallelopipeds,  having 
the  dimensions  a,  h,  c,  and  a',  V,  c',  respectively. 

To  prove  -^  =    ,       ,       , . 

Q       a'Xb'Xc' 

Let  i?  be  a  rectangular  parallelopiped-  having  the  dimen- 
sions a',  h',  and  c. 

Tlien  P  and  R  have  the  dimension  c  in  common. 
P        aXh 


Whence, 

±1       a  Xo 

And  B  and  (^  have  the  dimensions  a'  and  h'  in  common. 
Whence,  _  =  -. 

Multiplying  these  equations,  we  have 
P  ^   aXhXc 
Q       a'xh'Xc'' 


(§  498.) 
nmon. 
(§  496.) 


EXERCISES. 

1.  Two  rectangular  parallelopipeds  have  the  dimensions  6,  8, 
and  14,  and  7,  8,  and  9,  respectively.  What  is  the  ratio  of  their 
volumes  ? 

2.  Find  the  ratio  of  the  volumes  of  two  rectangular  parallelo- 
pipeds, whose  dimensions  are  8,  12,  and  21,  and  14,  15,  and  24, 
respectively. 

3 .  The  diagonals  of  a  rectangular  parallelopiped  are  equal. 


276 


SOLID   GEOMETRY.  —  BOOK   VII. 


Proposition  XI.     Theorem. 

500.  If  the  unit  of  volume  is  the  cube  ivhose  edge  is  the 
linear  unit,  the  volume  of  a  rectangular  parallelojjijjed  is 
equal  to  the  product  of  its  three  dimensions. 


/\ 

/ 

o 

i 

1 

L 

c 

A 

/ 

y — 

1 
/i 

/ 

X 

Let  a,  h,  and  c  be  the  dimensions  of  the  rectangnhir  i)ar- 
allelopiped  P  5  and  let  Q  be  the  unit  of  volume,  i.e.,  a  cube 
whose  edge  is  the  linear  unit. 

To  prove  Vol.  P  =  a  X  b  X  e. 

We  have        —  =  '^  X  ^  X  ^  =  a  X  b  X  c.  (§  499.) 

^1X1X1 
But  since  Q  is  the  unit  of  volume, 

^  =  vol.  p.  (§  469.) 


Whence, 


Q 

vol.  P  =  a  X  l>  X  c. 


501.  Cor.  I.  The  volume  of  a  cube  is  equal  to  the  cube  of 
its  edge. 

502.  Cor.  II.  If  c  be  taken  as. the  altitude  of  the  paral- 
lelopiped  P,  a  x  b  is  the  area  of  its  base.  (§  305.) 

Hence,  the  volume  of  a  rectangular  parallelopiped  is  equal 
to  the  product  of  its  base  and  altitude. 

503.  ScH.  I.  In  all  succeeding  theorems  relating  to  vol- 
umes, it  is  understood  that  the  unit  of  volume  is  the  cube 
whose  edge  is  the  linear  unit,  and  the  unit  of  surface  the 
square  whose  side  is  the  linear  unit.     (Compare  §  307.) 


PRISMS  AND   PARALLELOPirEDS.  277 

504.    Son.  II.    If    the    dimensions    of    the    rectangular 
parallelopiped  are  Ttiultiples  of  the  linear 
unit,  the  truth  of  Prop.  XI.  may  be  seen  /  /_./L/_y 


-/—/ 


-f-t- 

+t-l-|-- 

TtT-r 


by  dividing  the   solid   into   cubes,  each 
equal  to  the  unit  of  volume. 

Thus,  if  the  dimensions  of  the  rectan- 
gular parallelopiped  P  are  5  units,  4  units, 
and  3  units,  respectively,  the  solid  can  evi- 
dently be  divided  into  GO  cubes. 

In  this  case,  GO,  the  number  which  expresses  the  volume 
of  the  rectangular  parallelopiped,  is  the  product  of  5,  4,  and 
3,  the  numbers  which  express  the  lengths  of  its  edges. 


EXERCISES. 

4.  Find  the  altitude  of  a  rectangular  parallelopiped,  the  dimen- 
sions of  whose  base  are  21  and  30,  equivalent  to  a  rectangular  paral- 
lelopiped whose  dimensions  are  27,  28,  and  35. 

5.  Find  the  edge  of  a  cube  equivalent  to  a  rectangular  parallelo- 
piped whose  dimensions  are  9  in.,  1  ft.  9  in.,  and  4  ft.  1  in. 

6.  Find  the  volume,  and  the  area  of  the  entire  surface,  of  a  cube 
whose  edge  is  3^  in. 

7.  Find  the  area  of  the  entire  surface  of  a  rectangular  parallel- 
opiped, the  dimensions  of  whose  base  are  11  and  13,  and  volume  858. 

8.  Find  the  volume  of  a  rectangular  parallelopiped,  the  dimen- 
sions of  whose  base  are  14  and  9,  and  the  area  of  whose  entire 
surface  is  620. 

9.  Find  the  dimensions  of  the  base  of  a  rectangular  parallel- 
opiped, the  area  of  whose  entire  surface  is  320,  volume  336,  and 
altitude  4. 

10.  How  many  bricks,  each  8  in.  long,  2^  in.  wide,  and  2  in. 
thick,  will  be  required  to  build  a  wall  18  ft.  long,  3  ft.  high,  and 
11  in.  thick  ? 

11.  The  section  of  a  prism  made  by  a  plane  parallel  to  a  lateral 
edge  is  a  parallelogram. 

12.  The  square  of  a  diagonal  of  a  rectangular  parallelopiped  is 
equal  to  the  sum  of  the  squares  of  its  dimensions. 

13.  Find  the  length  of  the  diagonal  of  a  rectangular  parallelo- 
piped whose  dimensions  are  8,  9,  and  12. 


278 


SOLID   GEOMETKY.  —  BOOK   VII. 


Proposition  XII;     Theorem. 

505.    The  volume  of  any  ijarallelojpiijed   is  equal  to  the 
product  of  its  base  aiid  altitude. 


Let  AE  be  the  altitude  of  the  imrallelopiped  AC. 
To  prove  vol.  JC"  =  ABCD  X  AE. 

Produce  the  edges  AB,  AB',  jy  C ,  and  DC. 

On  AB  produced,  take  FG  =  AB ;  and  pass  the  i:>lanes 
FK'  and  GU'  perpendicular  to  FG,  forming  the  right  par- 
allelopiped  FH'. 

Then,  FB'  is  equivalent  to  AC.  '  (§  490.) 

Produce  the  edges  HG,  H' G\  K'F',  and  KF. 

On  HG  produced,  take  NM  =  JIG  ;  and  pass  the  planes 
NF'  and  ML'  perpendicular  to  N3I,  forming  the  right 
parallelopiped  LN\ 

Then,  LN'  is  equivalent  to  FB'.  (§  490.) 

Whence,  LN'  is  equivalent  to  A  C. 

Now  since  FG  is  perpendicular  to  the  plane  GB\  the 
planes  LB  and  MB'  are  perpendicular.  (§  443.) 

But  LMM'  is  the  plane  angle  of  the  diedral  LMBB' . 

(§  433.) 

Whence,  LMM'  is  a  right  angle.  (§  438.) 

Therefore,  LM'  is  a  rectangle,  and  LN'  is  a  rectangular 
parallelopiped. 


Whence, 


vol.  LN'  =  LMNP  X  MM'. 


(§  502.)        i 


PIIISMS   AND   PARALLELOPIPEDS.  279 

That  is,  vol.  AC  =  LMNP  X  MM'.  ( 1 ) 

But  the  rectangle  LMNP  is  equal  to  the  rectangle  i^G^^iC; 
for  they  have  equal  bases  MN  and  GH,  and  the  same  alti- 
tude. (§  113.) 

And  the  rectangle  FGHK  is  equivalent  to  the  parallelo- 
gram ABCD;  for  they  have  equal  bases  FG  and  AB,  and 
the  same  altitude.  (§  311.) 

Therefore,  LMNP  is  equivalent  to  ABCD. 

Again,  MM'  =  AE.  (§  422.) 

Substituting  these  values  in  (1),  we  have 
Yo\.  AC  =  ABCD  X  AE. 

'    Proposition  XIII.     Theorem. 

506.  The  volume  of  a  triangular  prism  is  equal  to  the 
product  of  its  base  and  altitude. 


Let  AE  be  the  altitude  of  the  triangular  prism  ABC-C. 
To  prove         vol.  ABC-C  =  ABC  X  AE. 

Construct  the  parallelopiped  ABCD-I/,  having  its  edges 
parallel  to  AB,  BC,  and  BB',  respectively. 

Then,  vol.  ABC-C  =  J-  vol.  ABCD-iy       (§  493.) 

=  ^ABCD  xAE  (§505.) 
=  ABC  X  AE.  (§  106.) 

EXERCISES. 

14.  Find  the  lateral  area  and  volume  of  a  regular  triangular 
prism,  each  side  of  whose  base  is  5,  and  whose  altitude  is  8. 

15.  The  diagonal  of  a  cube  is  equal  to  its  edge  multiplied  by  V3. 


280  SOLID   GEOMETRY.— BOOK   VII. 


PiiOPOSiTiON  XIV.     Theorem. 

507.    The  volume  of  any  prisTn  is  equal  to  the  product  of 
its  base  and  altitude. 


Any  prism  may  be  divided  into  triangular  prisms  by 
passing  planes  through  one  of  the  lateral  edges  and  the 
corresponding  diagonals  of  the  base. 

The  volume  of  each  triangular  prism  is  equal  to  the 
product  of  its  base  and  altitude  (§  506). 

Hence,  the  sum  of  the  volifines  of  the  triangular  prisms 
is  equal  to  the  sum  of  their  bases  multiplied  by  their  com- 
mon altitude. 

Therefore,  the  volume  of  the  given  prism  is  equal  to  the 
product  of  its  base  and  altitude. 

508.  CoR.  I.  Two  prisms  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

509.  Cor.  II.  1.  Two  prisms  having  equal  altitudes  are 
to  each  other  as  their  bases. 

2.  Tiao  prisms  having  equivalent  bases  are  to  each  other 
as  their  altitudes. 

3.  Any  two  prisms  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes. 


Ex.  16.    Find  tlio  lateral  area  and  volume  of  a  regular  hexagonal 
prism,  each  side  of  whose  base  is  3,  and  whose  altitude  is  9. 


PYKAMIDS. 


281 


PYRAMIDS. 


Definitions. 


510.  A  pyramid  is  a  polyedron  bounded  by  a  polygon, 
and  a  series  of  triangles  having  a  common 
vertex;  as  0-ABCDE. 

The  polygon  is  called  the  base  of  the 
pyramid,  and  the  common  vertex  of  the 
triangular  faces  is  called  the  vertex. 

The  triangular  faces  are  called  the  lat- 
eral faces,  and  their  intersections  the  lat- 
eral edges. 

The  sum  of  the  areas  of  the  lateral  faces  is  called  the 
lateral  area. 

511.  The  altitude  of  a  pyramid  is  the  perpendicular  dis- 
tance from  the  vertex  to  the  plane  of  the  base. 

512.  A  pyramid  is  called  triangular,  quadrangular,  etc., 
according  as  its  base  is  a  triangle,  quadrilateral,  etc. 

Note.     A  triangular  pyramid  is  a  tetraedron  (§  467). 

513.  A  regular  pyramid  is  a  pyramid 
whose  base  is  a  regular  polygon,  and  whose . 
vei-tex  lies  in  the  perpendicular  erected  at 
the  centre  of  the  base. 

514.  The  lateral  edges  of  a  regular  pyra- 
mid are  equal.  (§  407,  I.) 

Whence,  the  lateral  faces  are  equal  isosceles  triangles. 

(§  69.) 


515.  The  slant  height  of  a  regular  pyramid  is  the  per- 
pendicular distance  from  the  vertex  of  the  pyramid  to  any 
side  of  the  base. 

Or,  it  is  the  straight  line  drawn  from  the  vertex  of  the 
pyramid  to  the  middle  point  of  any  side  of  the  base.  (§  92.) 


282  SOLID   GEOMETRY.  —  BOOK  VII. 

516.  A  truncated  iJuramid  is  that  portion  of  a  pyramid 
included  between  the  base,  and  a  plane  cutting  all  the 
lateral  edges. 

517.  A  frustum  of  a  pyramid  is  a  truncated  pyramid 
whose  bases  are  parallel. 

The  altitude  of  the  frustum  is  the  per- 
pendicular distance  between  the  planes  .of 
its  bases. 

The  lateral  faces  of  a  frustum  of  a  pyr- 
amid are  trapezoids.  (§  417.) 

518.  The  lateral  faces  of  a  frustum  of  a  regular  pyramid 
are  equal. 

The  slant  height  of  a  frustum  of  a  regular  pyramid  is  the 
altitude  of  any  lateral  face. 

Proposition  XV.     Theorem. 

519.  If  a  pyramid  he  cut  hy  a  plane  parallel  to  its  base, 
I.    The  lateral  edges  and  the  altitude  are   divided  pro- 
portionally. 

II.    Tlie  section  is  similar  to  the  base. 


Let  A'C  be  a  plane  parallel  to  the  base  of  the  pyramid 
O-ABCB,  cutting  the  faces  OAB,  OBC,  OCD,  and  ODA 
in  the  lines  A'B',  B'C,  C'ly^  and  D'A',  and  the  altitude 
OP  at  P'. 


PYRAMIDS.  283 

T    m  OA'       OB'       OC    ^  OP' 

I.  To  prove.  = = etc.  =  — — . 

^  OA        OB        OC  OP 

Through  0  pass  the  plane  MN  parallel  to  ABCD. 

.p,_  OA'       OB'       OC    .  OP'        ,^  ,oK^ 

Then,  = = etc.  = .       (^  425.) 

OA        OB     •    OC  OP         ^^         ^ 

II.  To  i3rove  the  section  A'B'C'iy  similar  to  ABCD. 

We  have  A'B'  parallel  to  AB,  B'C  to  BC,  etc.      (§  417.) 
Then,  Z  A'B'C  =ZABC, 

Z.  B'C'B^  =  Z  BCD,  etc.  (§  424.) 

That  is,  the  polygons  A'B' CI/  and  ABCD  are  mutually 
equiangular. 

Again,  the  triangles  OAB'  and  OAB  are  similar.  (§  258.) 

^'>«"-'  -^  =  S-  (^) 

T    T1  OB'      B'C     ^ 

In  like  manner,  — —  =  — — - ,  etc. 

OB  B\y 

^"*'  "^^-if'"*"-  (§519,1.) 

wi.  ^^'    vc     c'ly    . 

Whence,  __  =  _.  =  ___,  etc. 

That  is,  the  polygons  A'B' CD'  and  ABCD  have  their 
homologous  sides  proportional. 

Therefore,  A'B' CD'  and  ABCD  are  similar.  (§  252.) 


520.    CoR.  I.   We  have 

area  A'B' CD'      A^ 


area  ABCD         j^b'^ 
But,  from  equation  (1)  of  §  519, 

A'B'  _  Of  ^  qp^ 

AB        OA        OP  ' 
area  A'B' CD'      'oF 


(§  323.) 
(§  519,  I.) 


Whence, 


area  ABCD         ^Qp^ 
That  is,  the  areas  of  tivo  parallel  sections  of  a  pyramid 
are  to  each  other  as  the  squares   of  their  distances  from 
the  vertex. 


284 


SOLID   GEOMETRY.  —  BOOK  VII. 


521.  Cor.  II.  If  two  pyramids  have  equal  altitudes  and 
equivalent  bases,  sections  parallel  to  their  bases  equally  dis- 
tant from  their  vertices  are  equivalent. 

Let  the  bases  of  the 
pyramids  0-ABC  and 
a-A'B'C  be  equivalent, 
and  let  the  altitude  of 
each  pyramid  be  H. 

Let  DEF  and  D'E'F' 
be  sections  parallel  to 
the  bases,  at  the  distance 
h  from  the  vertices. 

To  prove  DBF  equivalent  to  lYE'F'. 

^°"'  a^^^^m;  =  ^'  '^"'^  area  A'B'C  =  m'  ^'"''-^ 


Whence, 


area  BEF       area  B'E'F' 


area  ABC       area  A'B'C 
But  by  hypothesis,  area  ABC  =  area  A'B'C. 
Therefore,        area  DEF  =  area  B'E'F'. 


Proposition  XVI.     Theorem. 

522.    Two  triangular  pyram,ids  having  equal  altitudes  and 
equivilent  bases  are  equivalent. 


Let  o-abc  and  o'-ab'c'  be  two  triangular  pyramids  having 
equal  altitudes  and  equivalent  bases. 


PYRAMIDS.  285 

To  prove  vol.  o-abc  =  vol.  o'-a'h'c'. 

Place  the  pyramids  with  their  base^  in  the  same  plane, 
and  let  FQ  be  their  common  altitude. 

Divide  PQ  into  any  number  of  equal  parts ;  and  through 
the  points  of  division  pass  planes  parallel  to  the  plane  of 
the  bases,  cutting  o-dbe  in  the  sections  def  and  ghk,  and 
o'-a!h'c'  in  the  sections  d'e'f  and  g'h'k'. 

Then  def  is  equivalent  to  d'e'f,  and  ghk  to  g'h'k'.  (§  521.) 

With  abc,  def,  and  ghk  as  lower  bases,  construct  the 
prisms  A,  B,  and  C,  having  their  lateral  edges  equal  and 
parallel  to  ad;  and  with  d'e'f  and  g'h'k'  as  upper  bases, 
construct  the  prisms  B'  and  C,  having  their  lateral  edges 
equal  and  parallel  to  a'd'. 

Then,  the  prism  B  is  equivalent  to  B'.  (§  508.) 

In  like  manner,  C  is  equivalent  to  C 

Hence,  the  sum  of  the  prisms  circumscribed  about  o-abc 
exceeds  the  sum  of  the  prisms  inscribed  in  o'-a'h'c'  by  the 
prism  A. 

But  o-ahc  is  evidently  less  than  the  sum  of  the  prisms 
A,  B,  and  C ;  and  it  is  greater  than  the  sum  of  the  inscribed 
prisms,  equivalent  to  B'  and  C,  which  can  be  constructed 
with  def  and  ghk  as  upper  bases. 

Again,  o'-a'h'c'  is  greater  than  the  sum  of  the  prisms  B' 
and  C ',  and  it  is  less  than  the  sum  of  the  circumscribed 
prisms,  equivalent  to  A,  B,  and  C,  which  can  be  constructed 
with  a'h'c',  d'e'f,  and  g;h'k'  as  lower  bases. 

Hence,  the  difference  of  the  volumes  of  the  pyramids 
must  be  less  than  the  difference  of  the  volumes  of  the  two 
systems  of  prisms,  and  must  therefore  be  less  than  the 
volume  of  the  prism  A. 

Now  by  sufficiently  increasing  the  number  of  subdivisions 
of  BQ,  the  volume  of  the  prism  A.  may  be  made  less  than 
any  assigned  volume,  however  small. 

Therefore,  the  volumes  of  the  pyramids  cannot  differ  by 
any  volume,  however  small. 

Whence,  vol.  o-ahc  =  vol.  o'-a'h'c'. 


286 


SOLID  GEOMETRY. —BOOK  VII. 


Proposition  XVII.     Theorem. 

523.  The  lateral  area  of  a  regular  pyramid  is  equal  to 
the  perimeter  of  its  base  multiplied  by  one-half  its  slant 
/height. 


JjQt 'OH  be   the   slant   height   of   the    reguLar   pyramid 
0-ABCDE. 
To  prove 

lat.  area  0-ABCDE  =  (AB  +  BC  -\-  etc.)  X  i  Off. 

Now,  area  OAB  =  AB  X  ^  Off.  (§  313.) 

Also,  area  OBC  =  BC  X  k  Off;  etc.  (§  515.) 

Adding  these  equations,  we  have 

lat.  area  0-ABCDE  =  (AB  -{-  BC  +  etc.)  X  i  Off. 

524.    Cor.    The  lateral  ai^ea  of  a  frustum  of  a,  regular 
pyramid  is  equal  to  one-half  the  sum  of  , 

the  perim^eters  of  its  bases,  multiplied  by  A'^^--rT~~~-~yX>' 

its  slant  height.  /  l^fW's — v\ 

Let  HH'  be  the  slant  height  of  the     Ak:^j''f'' e'''''V--\b 
frustum  of  a  regular  pyramid  AD'.  -«\/ K 

To  prove 
lat.  area  AD^  =  ^  (AB'+  A'B'  +  BC  +  B'C  +  etc.)  X  Hff'. 

Now,  area  AA'B'B  =  |(J 7?  +  A'B')  x  ffff'.       (§  317.) 
Also,  area  BB'C'C  =-  \  {BC  +  B'C)  X  ffff'\  etc. 
Adding  these  equations,  we  have 
lat.  area  AD'  =  \(AB  +  A'B'  -\- BC  -{-  P/C  +  etc.)  X  HH'. 


d 


PYllAMlDS.  287 


Proposition  XVIII.     Theorem. 

525.    A  triangular  j^yramid  is  equivalent  to  one-third  of  a 
triangular  prism  having  the  same  base  and  altitude. 


Let  0-ABC  be  a  triangular  pyramid. 
Upon  the  base  ABC,  construct   the    prism  ABC-ODE, 
having  its  lateral  edges  equal  and  parallel  to  OB. 
To  prove    vol.  0-4BC  =  ^  vol.  ABC-ODE. 

The  prism  ABC-ODE  is  composed  of  the  triangular 
pyramid  0-ABC,  and  the  quadrangular  pyramid  0-ACDE. 

Divide  0-ACDE  into  two  triangular  pyramids,  0-ACE 
and  0-CDE,  by  passing  a  plane  through  0,  C,  and  E. 

Now,  0-ACE  and  0-CDE  have  the  same  altitude. 

And  since  CE  is  a  diagonal  of  the  parallelogram  ACDE 
they  have  equal  bases,  ACE  and  CDE.  (§  106.) 

Hence,  yo\.  0-ACE  =  yo\.  0-CDE.  (§522.) 

Again,  the  pyramid  0-CDE  may  be  regarded  as  having 
its  vertex  at  C,  and  the  triangle  ODE  for  its  base. 

Then,  the  pyramids  0-ABC  and  C-ODE  have  the  same 
altitude.  (§  422.) 

They  have  also  equal  bases,  ABC  and  ODE.  (§  472.) 

Hence,  ^         vol.  0-ABC  =  vol.  C-ODE.  (§  522.) 

Then,  vol.  0-ABC  =:yo\.  0-ACE  =  vol.  0-CDE. 

Whence,     vol.  0-ABC  =  ^  vol.  ABC-ODE. 

526.  Cor.  The  volume  of  a  triangular  pyramid  is  equal 
to  one-third  the  product  of  its  base  and  altitude.  (§  506.) 


288  SOLID    GEOMETKY.  —  BOOK   VII. 


Proposition  XIX.     Theorem. 

527.    The  volume  of  any  pyramid  is  equal  to  one-third  the 
product  of  its  base  and  altitude. 


Any  pyramid  may  be  divided  into  triangular  pyramids 
by  passing  planes  tlirough  one  of  the  lateral  edges  and  the 
corresponding  diagonals  of  the  base. 

The  volume  of  each  triangular  pyramid  is  equal  to  one- 
third  the  product  of  its  base  and  altitude  (§  526). 

Hence,  the  sum  of  the  volumes  of  the  triangular  pyra- 
mids is  equal  to  the  sum  of  their  bases  multiplied  by 
one-third  their  common  altitude. 

Therefore,  the  volume  of  the  given  pyramid  is  equal  to 
one-third  the  product  of  its  base  and  altitude. 

528.  Cor.  I.  Two  pyramids  having  equivalent  bases  and 
equal  altitudes  are  equivalent. 

529.  Cor.  II.  1.  Two  pyramids  having  equal  altitudes 
are  to  each  other  as  their  bases. 

2.  Two  pyramids  having  equivalent  bases  are  to  each  other 
as  their  altitudes. 

3.  Any  two  pyramids  are  to  each  other  as  the  products  of 
their  bases  by  their  altitudes. 

530.  ScH.  The  volume  of  any  polyedron  may  be  obtained 
by  dividing  it  into  pyramids. 


PYRAMIDS. 


289 


Proposition  XX.     Theorem. 

531.  Two  tetraedrons  having  a  triedral  of  one  equal  to 
a  triedral  of  the  other,  are  to  each  other  as  the  products  of 
the  edges  including  the  equal  triedrals. 


Let  V  and  F'  denote  the   volumes   of   tlie  tetraedrons 
0-ABC  and  O-A'B'C,  having  the  common  triedral  0. 
V_  ^    OA  X  OB  X  PC 
V 


To  prove 


OA'  X  OB'  X  OC 
Draw  CP  and  CJ'P'  perpendicular  to  the  face  OA'B\ 
Let  their  plane  intersect  OA'B'  in  the  line  OPP'. 
Now,  OAB  and  OA'B'  are  the  bases,  and  CP  and  CP' 
the   altitudes,    of    the    triangular    pyramids    C-OAB    and 
C'-OA'B'. 

V_  _   OAB  X  CP 
V 


Whence, 


OA'B'  X  CP' 


OAB 


X 


CP 


But, 


OA'B'       CP' 
OAB         OA  X  OB 


(§  529,  3.) 

(1) 

(§  322.) 


OA'B'~      OA'  X  OB' 
Also,  the  right  triangles  OCP  and  OCP'  are  similar. 

(§  257.) 
Whence, 


CP 

CP' 


OC 
OC 


Substituting  these  values  in  (1),  we  have 


V^ 
V 


OA  XOB    ,,  OC 


OAxOB  X  OC 


OA'  X  OB'       OC       OA'  X  OB'  x  OC 


290  SOLID   GEOMETRY.  —  BOOK    VII. 

Proposition  XXI.     Theorem. 

532.  The  volume  of  a  frustum  of  a  pyramM  is  equal  to 
the  sum  of  its  bases  and  a  mean  proportional  between  its 
bases,  multiplied  by  one-third  its  altitude. 


Let  AD'  be  a  frustum  of  any  pyramid.  0-ABCDE. 
Denote  the  area  of  the  lower  base  by  B,  the  area  of  the 
upper  base  by  b,  and  the  altitude  by  H. 

To  prove  vol.  AD'  =  {B  +  b-\-  ^B  X  b)  x  ^  ff.    (§  232.) 

Draw  the  altitude  OP,  cutting  A'B'C'D'E'  at  P'. 

Now,  vol.  AD'  =  vol.  0-ABCDE  —  vol.  0-A'B'C'D'E' 
=  BxlOP-bx^OP'  (§  527.) 

=  Bxh{H  ^  OP')  -bx^OP' 
'.^    =B  X^jH  +  BX^OP'-bX^OP' 
=  B  X  iH-i-(B-b)  X  ^  OP'. 

But,  B:b=  op'  '  OP'^ . 

Taking  the  square  root  of  each  term,  we  have 
V^:  Vb=  OP:  OP'.. 

Then,  V^-  V^ :  V^*  =  OP  -  OP'  or  // : 

Whence,.     (V^—  V^)  X  OP'  =  ^b  X  H. 
Multiplying  both  members  by  V^  +  V^, 

(B-b)X  OP'  =  (VBxb  -{-b)xIT. 
Substituting  in  (1),  We  have 

vol.  AD'  =  B  X  ^H  ^  {-^/WxJ)  +  b)X\H 
^(^B-irb-Y  -VBXb)  X  I  H, 


(§  520.) 

(§  242.) 

OP'. 

(§  237.) 

(§  231.) 

rYllAMlDS.  291 


Proposition   XXII.     Theorem. 

533.  The  volume  of  a  truncated  triangular  prism  is  equal 
to  the  product  of  a  right  section  bij  one-third  the  sum  of  its 
lateral  edges. 


F 


Let  GHC  and  DKL  be  right  sections  of  the  truncated 
triangular  prism  ABC-DEF. 
To  prove 

voL  ABC-DEF  =  GHC  X  h  {^D  +  BE  +  CF). 

Draw  DM  perpendicular  to  KL. 

The  truncated  prism  is  composed  of  the  triangular  prism 
GHC-DKL,  and  the  pyramids  D-EKLF  and  C-ABHG. 

Now  since  the  lateral  edges  of  a  prism  are  equal, 
vol.  GHC-DKL  =  GHC  X  GD  (§  506.) 

=  GHC  X  i  (GD  +  HK-\-  CL).         (1) 
Again,  DM  is  the  altitude  of  the  pyramid  D-EKLF. 

(§  440.) 
Whence,    vol.  D-EKLF  =  EKLF  X  ^  DM         (§  527.) 
=  i  (KE  +  LF)  XKL  XI  DM  (§  317.) 

=  iKL  xDMxi  (KE+LF). 
But,       1  KL  X  D3I  =  area  DKL  =  area  GHC.     (§  313.) 
Hence,  vol.  D-EKLF  =  GHC  X  h  {KE  +  LF).  ( 2 ) 

In  like  manner,  we  may  prove  ^ 

vol.  C-ABHG  =  GHC  X  ^  {AG  +  BH).         (3) 
Adding  (1),  (2),  and  (3),  we  have 
vol.  ABC-DEF 
=  GHCx^{AG-\-  GD  +  BH  -\-  HK  +  KE  +  CL-\-LF) 
'  ==  GHC  X  i  {^D  +  J^^  +  CF). 


292 


SOLID   GEOMETRY.  —  BOOK   YII. 


534.  Cor.  The  volume  of  a  truncated  rujht  triangular 
prisin  is  equal  to  the  product  of  its  hase  by  one-tldrd  the  sum 
of  its  lateral  edges. 

EXERCISES. 

17.  Each  side  of  the  base  of  a  regular  triangular  pyramid  is  6, 
and  its  altitude  is  4.     Find  its  lateral  edge,  lateral  area,  and  volume. 

Let  D  be  the  centre  of  the  base  of  the  regular 
triangular  pyramid  0-ABC,  and  draw  OD  and 
AD;  also,  draw  CBE  perpendicular  to  AB,  and 
join  OE. 

Then,  lat.  edge  OA 

V28: 

■v/19. 


Now,  AD  =  AB 


=  Voif  +  AD^=  Vl6  +  12 
The  slant  ht.  OE 


2Vl. 


V 

Then,  lat.  area  0-ABC 
Again,  CE 


^B& 


6A^  -  AE^  =  V28 

9  Vl9"(§523). 

VS6 


BE"  =  Vse  -  9  =  \/27  =  3  V3. 
Then,  area  ABC  =  ^  x  6  x  3  V3  (§  313)  =  9  Vs. 
Whence,  vol.  0-ABC  =  ix9>/3x4(§  526)  =  12  V3. 

18.  Find  the  lateral  edge,  lateral  area,  and  volume  of  a  fnistum 
of  a  regular  quadrangular  pyramid,  the  sides  of  whose  bases  are  17 
and  7,  and  whose  altitude  is  12. 

Let  O  and  O'  be  the  centres  of  the  bases 
of  the  frustum  of  a  regular  quadrangular 
pyramid  AC. 

Draw  OE  and  O'E'  perpendicular  to  AB 
and  A'B\  and  E'F  and  A'G  perpendicular 
to  OE  and  AB;  also,  draw  00'  and  EE\ 

Now,  EF=  OE-  O'E'  =  8^  -  31  =  5. 

.^EF^  +  W^^  =  \/25  +  144  = 


Then,  slant  ht.  EE 

Whence,  lat.  area  AC  =  -J  (68  +  28)  x  13  (§524)  : 

Again,  AG  =  AE-  A'E'  =  8i  —  3i  =  5;  and  A'G 

Whence,  lat.  edge  AA'  =  VaG^  +  AJG^  =  V25  +  169 
Again,  area  AC  =  11^  =  289,  and  area  A'C  =  1^  =  49. 
Then,  a  mean  proportional  between  the  areas  of  the  bases 

^  VlV  xr'  =  l7  Xl=  119. 
Whence,  vol.  AC  =  (289  +  49  +  119)  X  4  (§  532)  =  1828. 


169  =  13. 


PYRAMIDS.  293 

Find  the  lateral  edge,  lateral  area,  and  volume 

19.  Of  a  regular  triangular  pyramid,  eacl*  side  of  whose  base  is 
12,  and  whose  altitude  is  15. 

20.  Of  a  regular  quadrangular  pyramid,  each  side  of  whose  base 
is  3,  and  whose  altitude  is  5. 

21.  Of  a  regular  hexagonal  pyramid,  each  side  of  whose  base  is 
4,  and  whose  altitude  is  9. 

22.  Of  a  frustum  of  a  regular  triangular  pyramid,  the  sides  of 
whose  bases  are  18  and  G,  and  whose  altitude  is  24. 

23.  Of  a  frustum  of  a  regular  quadrangular  pyramid,  the  sides  of 
whose  bases  are  9  and  5,  and  whose  altitude  is  10. 

24.  Of  a  frustum  of  a  regular  hexagonal  pyramid,  the  sides  of 
whose  bases  are  8  and  4,  and  whose  altitude  is  12. 

25.  Find  the  volume  of  a  truncated  right  triangular  prism,  the 
sides  of  whose  base  are  5,  12,  and  13,  and  whose  lateral  edges  are 
3,  7,  and  5,  respectively. 

26.  Find  the  volume  of  a  truncated  regular  quadrangular  prism, 
each  side  of  whose  base  is  8,  and  whose  lateral  edges,  taken  in 
order,  are  2,  6,  8,  and  4,  respectively. 

27.  The  volume  of  a  cube  is  4|f  cu.  ft.  What  is  the  area  of  its 
entire  surface  in  square  inches  ? 

28.  A  box  made  of  2  in.  plank,  without  a  cover,  measures  on  the 
outside  3  ft.  2  in.  long,  2  ft.  3  in.  wide,  and  1  ft.  6  in.  deep.  How 
many  cubic  feet  of  material  were  used  in  its  construction  ? 

29.  The  volume  of  a  right  prism  is  2310,  and  its  base  is  a  right 
triangle  whose  legs  are  20  and  21.     Find  its  lateral  area. 

30.  Find  the  lateral  area  and  volume  of  a  right  triangular  prism, 
the  sides  of  whose  base  are  4,  7,  and  9,  and  whose  altitude  is  8. 

31.  Find  the  volume  of  a  truncated  right  triangular  prism,  whose 
lateral  edges  are  11,  14,  and  17,  having  for  its  base  an  isosceles  tri- 
angle whose  sides  are  10,  13,  and  13. 

32.  The  altitude  of  a  pyramid  is  12  in.,  and  its  base  is  a  square 
9  in.  on  a  side.  What  is  the  area  of  a  section  parallel  to  the  base, 
whose  distance  from  the  vertex  is  8  in.? 

33.  The  altitude  of  a  pyramid  is  20  in.,  and  its  base  is  a  rectangle 
whose  dimensions  are  10  in.  and  1.5  in.  What  is  the  distance  from 
the  vertex  of  a  section  parallel  to  the  base,  whose  area  is  54  sq.  in.? 


294  SOLID   GEOMETRY.  —  BOOK  YII. 

34.  The  diagonal  of  a  cube  is  8  VE.  Find  its  volume,  and  the  area 
of  its  entire  surface. 

35.  A  trench  is  124  ft.  long,  2^  ft.  deep,  6  ft.  wide  at  the  top,  and 
5  ft.  wide  at  the  bottom.  How  many  cubic  feet  of  water  will  it 
contain  ?  ^ 

36.  The  volume  of  a  regular  triangular  prism  is  96  VS,  and  one 
side  of  its  base  is  8.     Find  its  lateral  area. 

37.  The  lateral  area  and  volume  of  a  regular  hexagonal  prism  are 
60  and  15  V3,  respectively.   Find  its  altitude,  and  one  side  of  its  base. 

38.  The  slant  height  and  lateral  edge  of  a  regular  quadrangular 
pyramid  are  25  and  V674,  respectively.  Find  its  lateral  area  and 
volume. 

39.  The  altitude  and  slant  height  of  a  regular  hexagonal  pyramid 
are  15  and  17,  respectively.     Find  its  lateral  edge  and  volume. 

40.  The  lateral  edge  of  a  frustum  of  a  regular  hexagonal  pyramid 
is  10,  and  the  sides  of  its  bases  are  10  and  4,  respectively.  Find  its 
lateral  area  and  volume. 

41.  Find  the  lateral  area  and  volume  of  a  regular  quadrangular 
pyramid,  the  area  of  whose  base  is  100,  and  whose  lateral  edge  is  13. 

42.  Find  the  lateral  area  and  volume  of  a  frustum  of  a  regular 
triangular  pyramid,  the  sides  of  whose  bases  are  12  and  6,  and  whose 
lateral  edge  is  5. 

43.  The  lateral  edges  of  a  frustum  of  a  quadrangular  pyramid 
are  equal;  and  its  bases  are  rectangles,  whose  sides  are  27  and  15, 
and  9  and  5,  respectively.  If  the  altitude  of  the  frustum  is  12,  find 
its  lateral  area  and  volume. 

44.  Any  straight  line  drawn  through  the  centre  of  a  parallelo- 
piped,  terminating  in  a  pair  of  oi)i)osite  faces,  is  bisected  at  that 
point. 

45.  The  lateral  surface  of  a  pyramid  is  greater  than  its  base. 

46.  The  volume  of  a  regular  prism  is  equal  to  its  lateral  area, 
multiplied  by  one-half  the  apothem  of  its  base. 

47.  The  volume  of  a  regular  pyramid  is  equal  to  its  lateral  area, 
multiplied  by  one-third  tha  distance  from  the  centre  of  its  base  to  any 
lateral  face. 

48.  If  E,  F,  G,  and  II  are  the  middle  points  of  the  edges  AB, 
AD,  CD,  and  BC,  respectively,  of  the  tetraedron  ABCD,  prove  that 
EFGII  is  a  parallelogram. 


PYRAMIDS.  295 

49.  Find  the  area  of  the  base  of  a  regular  quadrangular  pyramid, 
whose  lateral  faces  are  equilateral  triangles,  and  whose  altitude  is  5. 

50.  Two  tetraedrons  are  equal  if  two  faces  and  the  included 
diedral  of  one  are  equal,  respectively,  to  two  faces  and  the  included 
diedral  of  the  other,  if  the  equal  parts  are  similarly  placed. 

51.  Two  tetraedrons  are  equal  if  three  faces  of  one  are  equal,  re- 
spectively, to  three  faces  of  the  other,  if  the  equal  parts  are  similarly 
placed. 

52.  Find  the  area  of  the  entire  surface  and  the  volume  of  a  trian- 
gular pyramid,  each  of  whose  edges  is  2. 

53.  The  areas  of  the  bases  of  a  frustum  of  a  pyramid  are  12  and 
75,  and  its  altitude  is  9.     What  is  the  altitude  of  the  pyramid  ? 

54.  The  sum  of  two  opposite  lateral  edges  of  a  truncated  paral- 
lelopiped  is  equal  to  the  smn  of  the  other  two  lateral  edges. 

55.  The  volume  of  a  truncated  parallelopiped  is  equal  to  the 
area  of  a  right  section,  multiplied  by  one-fourth  the  sum  of  the  lat- 
eral edges. 

56.  A  plane  passed  through  the  centre  of  a  parallelopiped  divided 
it  into  two  equivalent  solids. 

57.  If  ABCD  is  a  rectangle,  and  EF  any  straight  line  not  in  its 
plane  parallel  to  AB^  the  volume  of  the  solid  bounded  by  the  figures 
ABCD,  ABFE,  CDEF,  ABE,  and  BCF,  is  equal  to 

\hx  ADx{2AB  +  EF), 
where  h  is  the  perpendicular  from  E  to  ABCD.     (§  533.) 

58.  If  ABCD  and  EFGII  are  rectangles  lying  in  parallel  planes, 
AB  and  BC  being  j)arallel  to  EF  and  EG,  respectively,  the  solid 
bounded  by  the  figures  ABCD,  EFGR,  ABFE,  BCGF,  CDHG,  and 
DAEII,  is  called  a  rectangular  prismoid. 

ABCD  and  EFGII  are  called  the  bases  of  the  rectangular  prismoid, 
and  the  perpendicular  distance  between  them  the  altitude. 

Prove  that  the  volume  of  a  rectangular  prismoid  is  equal  to  the 
sum  of  its  bases,  plus  four  times  a  section  equidistant  from  the  bases, 
multiplied  by  one-sixth  the  altitude.     (Ex.  57.) 

59.  Find  the  volume  of  a  rectangular  prismoid,  the  sides  of  whose 
bases  are  10  and  7,  and  6  and  5,  respectively,  and  whose  altitude  is  9. 

60.  The  volume  of  a  triangular  prism  is  equal  to  a  lateral  face, 
multiplied  by  one-half  its  perpendicular  distance  from  any  point  in 
the  opposite  lateral  edge. 


296  SOLID   GEOMETRY.  — BOOK  VII. 

61.  The  volume  of  a  truncated  right  parallelopiped  is  equal  to  the 
area  of  its  lower  base,  multiplied  by  the  perpendicular  drawn  to 
the  lower  base  from  the  centre  of  the  upper  base. 

62.  The  perpendicular  drawn  to  the  lower  base  of  a  truncated 
right  triangular  prism  from  the  intersection  of  the  medians  of  the 
upper  base,  is  equal  to  one-third  the  sura  of  the  lateral  edges. 

63.  The  three  planes  passing  through  the  lateral  edges  of  a  tri- 
angular pyramid,  bisecting  the  sides  of  the  base,  meet  in  a  common 
straight  line. 

64.  A  frustum  of  any  pyramid  is  equivalent  to  the  sum  of  three 
pyramids,  having  for  their  common  altitude  the  altitude  of  the  frus- 
tum, and  for  their  bases  the  lower  base,  the  upper  base,  and  a  mean 
proportional  between  the  bases,  of  the  frustum. 

65.  A  monument  is  in  the  form  of  a  frustum  of  a  regular  quad- 
rangular pyramid  8  ft.  in  height,  the  sides  of  whose  bases  are  3  ft.  and 
2  ft.,  respectively,  surmounted  by  a  regular  quadrangular  pyramid 
2  ft.  in  height.     What  is  its  weight,  at  180  lb.  to  the  cubic  foot  ? 

66.  The  altitude  and  lateral  edge  of  a  frustum  of  a  regular  tri- 
angular pyramid  are  8  and  10,  respectively,  and  each  side  of  its  upper 
base  is  2  Vs.     Find  its  volume  and  lateral  area. 

67.  A  railway  embankment,  1620  ft.  in  length,  is  8^  ft.  wide  at  the 
top,  21^  ft.  wide  at  the  bottom,  and  6  ft.  4  in.  high.  How  many 
cubic  yards  of  earthwork  does  it  contain  ? 

68.  The  sides  of  the  base,  AB,  BC,  and  CA,  of  a  truncated  right 
triangular  prism  ABC-DEF,  are  15,  4,  and  12,  respectively,  and  the 
lateral  edges,  AD,  BE,  and  CF,  are  15,  7,  and  10,  respectively.  Find 
the  area  of  the  upper  base,  BEF. 

69.  If  ABCD  is  a  tetraedron,  the  section  made  by  a  plane  par- 
allel to  each  of  the  edges  AB  and  CD  is  a  parallelogram.     (§  415.) 

70.  In  a  tetraedron  ABCD,  a  plane  is  drawn  through  the  edge  CD 
perpendicular  to  AB,  intersecting  the  faces  ABC  and  ABD  in  CE 
and  ED.     If  the  bisector  of  the  angle  CED  meets  CD  at  F,  prove 

CFiDF  =  area  ABC  :  area  ABD. 

71.  The  sum  of  the  squares  of  the  four  diagonals  of  any  paral- 
lelopiped is  equal  to  the  sum  of  the  squares  of  its  twelve  edges. 
(Ex.  75,  p.  226.) 

72.  If  the  four  diagonals  of  a  quadrangular  prism  pass  through  a 
common  point,  the  prism  is  a  parallelopiped. 


SIMILAR  POLYEDRONS  297 


SIMILAR  POLYEDRONS. 


535.  Def.  Two  polyedrons  are  said  to  be  similar  when 
they  are  bounded  by  the  same  number  of  faces,  similar 
each  to  each  and  simihirly  placed,  and  have  their  homol- 
"ogous  polyedrals  equal. 


Proposition   XXIIT.     Theorem. 

536.  The  ratio  of  any  tiuo  homologous  edges  of  two  similar 
polyedrons  is  e(j[ual  to  the  ratio  of  any  other  two  homologous 
edges. 


In  the  similar  polyedrons  AF  and  AF'^  let  the  edges  AB 
and  EF  be  homologous  to  the  edges  AB'  and  E'F'. 

^  AB        EF 

lo  prove  -— —  =  — — -  . 

^  A!B'      E'F' 

The  face  J  C  is  similar  to  A'C,  and  BF  to  jyF',    (§  535.) 
Whence,  ^f  =  ||.  =  ^  •  (§253,1.) 


537.   CoR.  I.   We  have 

area  ABCD  AB^ 


3ivesi  A' B'C'jy      AW 


(§  323.) 


(§  536.) 


E'F''' 

Therefore,  any  tivo  homologous  faces  of  two  similar  poly- 
edrons are  to  each  other  as  the  squares  of  any  two  homolo- 
gous edges. 


^9^  SOLID   GEOMETRY.  —  BOOK  Yll. 

538.   Cor.  II.   We  have 

ABCD  CDEF  ^  Ib" 


=  etc.  =  ^^.    (§537.) 


A'B'G'iy       C'D'E'F'  A^' 

Whence, 

ABCD  4-  CDEF  +  etc.       _  AB^ 


A'B'C'D'  +  C'D'E'F'  +  etc.       a'B'''  ' 

Hence,  the  entire  surfaces  of  two  similar  polyedrons  ore  to 
each  other  as  the  squares  of  any  two  homologous  edges. 

Proposition  XXIV.     Theorem. 

539.  Two  tetraedrons  are  similar  when  the  faces  includ- 
ing a  triedral  of  one  are  similar  to  the  faces  including  a  tri- 
edral  of  the  other,  and  similarly  placed. 


In  the  tetraedrons  ABCD  and  A'B'C'D',  let  the  faces 
ABC,  ACD,  and  ADB  be  similar  to  the  faces  A'B'C, 
A' CD',  and  A'D'B',  respectively. 

To  prove  ABCD  and  A'B'C'D'  similar. 

From  the  given  similar  faces,  we  have 

BC   _  AC_  _    CD^  _  AD^  _  BD_ 
B'C       A'C       CD'      A'D'      B'D'' 
Hence,  the  faces  BCD  and  B'C'D'  are  similar.      (§  259.) 
Again,  since  the  angles  BAC,  CAD,  and  DAB  are  equal 
respectively  to  the  angles^'^'C",  CA'D',  and  D'A'B',  the 
triedrals  A-BCD  and  A'-B'CD'  are  equal.       -       (§  464,  I.) 
In  like  manner,  any  two  homologous  triedrals  are  equal. 
Hence,  ABCD  and  A'B'Ciy  are  similar.  (§  535.) 


SIMILAR  POLYEDRONS.  299 

540.  Cor.  If  a  tetraedron  he  cut  by  a  jilane  parallel  io 
one  of  its  faces,  the  tetraedron  cut  off  is  similar  to  the  giveri 
tetraedron. 

Proposition  XXV.     Theorem. 

541.  Two  tetraedrons  are  similar  when  a  diedral  of  one  is 
eq^iial  to  a  diedral  of  the  other,  and  the  faces  including  the 
equal  diedrals  are  similar  each  to  each,  and  similarly  placed. 


In  the  tetraedrons  ABCB  and  A'B'C'iy,  let  the  diedral 
AB  be  equal  to  A'B' ;  and  let  the  faces  ABC  and  ABD  be 
similar  to  A'B'C  and  AB'B',  respectively. 

To  prove  ABCB  and  AB'C'iy  similar.' 

Let  the  tetraedron  A'B'C'jy  be  applied  to  ABCD,  so  that 
the  diedral  AB'  shall  coincide  with  its  equal  AB,  the  point 
A!  falling  at  ^. 

Then  since  Z  B'A'C  =  ABAC,  and  Z  B'AIf  =  Z  BAD, 
the  edge  AC  will  coincide  with  AC,  and  AB^  with  AB. 

Therefore,  Z  C'AB^  =  Z  CAB. 

Again,  from  the  given  similar  faces,  we  have 

AC        AB        AB  ' 
Hence,  the  triangle  CAB'  is  similar  to  CAB.       (§  260.) 
Then,  the  faces  including  the  triedral  A-B'C'B'  are  simi- 
lar to  the  faces  including  the  triedral  A-BCB,  and  similarly 
placed. 

Therefore,  ABCB  and  AB'C'B'  are  similar.  (§  539.) 


300  SOLID   GEOMETRY.  —  BOOK   VII. 

Proposition  XXVI.     Theorem. 

542.  Tivo  similar  polyedrons  may  he  decomjjosed  into  the 
same  number  of  tetraedrons,  similar  each  to  eacJi^  and  simi- 
larly placed. 


Let  AF  and  A'F'  be  two  similar  polyedrons,  the  vertices 
A  and  A'  being  homologous. 

To  prove  that  they  may  be  decomposed  into  the  same  num- 
ber of  tetraedrons,  similar  each  to  each,  and  similarly  placed. 

Divide  all  the  faces  of  AF,  except  those  having  J  as  a 
vertex,  into  triangles ;  and  draw  straight  lines  from  A  to 
their  vertices. 

In  like  manner,  divide  all  the  faces  of  A'F',  except  those 
having  A'  as  a  vertex,  into  triangles  similar  to  those  in  AF, 
and  similarly  placed.  (§  267.) 

Draw  straight  lines  from  A'  to  their  vertices. 

The  given  polyedrons  are  then  decomposed  into  the  same 
number  of  tetraedrons,  similarly  placed. 

Let  ABCF  and  AB'C'F'  be  two  homologous  tetraedrons. 

The  triangles  ABC  and  BCF  are  similar  to  A'B'C  and 
B'C'F',  respectively.  (§  267.) 

And  since  the  given  polyedrons  are  similar,  the  homolo- 
gous diedrals  BC  and  B'C  are  equal. 

Therefore,  ABCF  and  A' B'C'F'  are  similar.  (§  541.) 

In  like  manner,  we  may  prove  any  two  homologous  tetrae- 
drons similar. 

Hence,  the  given  polyedrons  are  decomposed  into  the 
same  number  of  tetraedrons,  similar  each  to  each,  and 
similarly  placed. 


SIMILAR  POLYEDRONS.  301 

Proposition  XXVII.     Theorem. 

543.    Two  similar  tetraedrons  are  to  each  other  as  the 
cubes  of  their  homologous  edges. 


Let  V  and  V  denote  the  volumes  of  the  similar  tetra- 
edrons ABCD  and  A' BCD',  the  vertices  A  and  A'  being 
homologous. 

To  prove 


Since  the  triedrals  at  A  and  A'  are  equal,  we  have 

V_^     ABxACxAD 

V       A'B'  X  A'C  X  A'ly  (§  ^^^•) 

AB        AC        AD 

^  1777}  ^  -JTRr  K^) 


A'B'       A'C       A'D' 
-P   .  AC        AD        AB  ,.  ^_. 

^^*'  ^C'^AD  =  A^''  (^^^^-^ 

Substituting  in  (1),  we  have 

V         ^B    X  ^^  X   ^^         ^^ 


V       A'B'       A'B'       A'B'       j^^ 

544.  Cor.  Any  two  similar  polyedrons  are  to  each  other 
as  the  cubes  of  their  homologous  edges. 

For  any  two  similar  polyedrons  may  be  decomposed  into 
the  same  number  of  tetraedrons,  similar  each  to  each  (§  542). 

And  any  two  homologous  tetraedrons  are  to  each  other  as 
the  cubes  of  their  homologous  edges  (§  543),  or  as  the  cubes 
of  any  two  homologous  edges  of  the  polyedrons  (§  536). 


302        SOLID  GEOMETRY.  —  BOOK  VII. 

REGULAR  POLYEDRONS. 

545.  Def.  a  regular  polyedron  is  a  polyedron  whose 
faces  are  equal  regular  polygons,  and  whose  polyedrals  are 
all  equal. 

Proposition  XXVIII.     Theorem. 

546.  Not  TYiore  than  five  regular  convex  pohjedrons  are 
possible. 

A  convex  polyedral  must  have  at  least  three  faces,  and 
the  sum  of  its  face  angles  must  be  less  than  360°  (§  463). 

1.  With  equilateral  triangles. 

Since  the  angle  of  an  equilateral  triangle  is  60°,  we  may 
form  a  convex  polyedral  by  combining  either  3,  4,  or  5  equi- 
lateral triangles. 

Not  more  than  5  equilateral  triangles  can  be  combined  to 
form  a  convex  polyedral.  (§  463.) 

Hence,  not  more  than  three  regular  convex  polyedrons 
can  be  formed  with  equilateral  triangles. 

2.  With  squares. 

Since  the  angle  of  a  square  is  90°,  we  may  form  a  convex 
polyedral  by  combining  3  squares. 

Not  more  than  3  squares  can  be  combined  to  form  a  con- 
vex polyedral. 

Hence,  not  more  than  one  regular  convex  polyedron  can 
be  formed  with  squares. 

3.  With  regular  pentagons. 

Since  the  angle  of  a  regular  pentagon  is  108°,  we  may 
form  a  convex  polyedral  by  combining  3  regular  pentagons. 

Not  more  than  3  regular  pentagons  can  be  combined  to 
form  a  convex  polyedral. 

Hence,  not  more  than  one  regular  convex  polyedron  can 
be  formed  with  regular  pentagons. 

Since  the  angle  of  a  regular  hexagon  is  120°,  no  convex 
polyedral  can  be  formed  by  combining  regular  hexagons. 


REGULAR  POLYEDRONS.  303 

111  like  manner,  no  convex  polyedral  can  be  formed  by 
combining  regular  polygons  of  more  than  six  sides. 

Hence,  not  more  than  five  regular  convex  polyedrons  are 
possible. 

Proposition  XXIX.     Problem. 

547.    With  a  given  edge,  to  construct  a  regular  polyedron. 
We  will  now  prove,  by  actual  construction,  that  five  regu- 
lar convex  polyedrons  are  possible : 

1.  The  regular  tetraedron,  bounded  by  4  equilateral  tri- 
angles. 

2.  The  regular  hexaedron,  or  cube,  bounded  by  6  squares. 

3.  The  regular  octaedron,  bounded  by  8  equilateral  tri- 
angles. 

4.  The  regular  dodecaedron,  bounded  by  12  regular  pen- 
tagons. 

5.  The  regular  icosaedron,  bounded  by  20  equilateral  tri- 
angles. 

1.    To  construct  a  regular  tetraedron.  a 

Let  AB  be  the  given  edge.  / 

Construct  the  equilateral  triangle  AB  C.  / 

At  its  centre  E,  draw  ED  perpendicu-         / 
lar  to  ABC',  and  take  the  point  D  so  that        '^<;^ 
AD  =  AB.  "^ 

Draw  AD,  BD,  and  CD. 
Then,  ABCD  is  a  regular  tetraedron. 
For  since  A,  B,  and  C  are  equally  distant  from  E, 

AD  =  BD=  CD.  (§  407,  I.) 

Hence,  the  six  edges  of  the  tetraedron  are  all  equal. 
Then,  the  faces  are  equal  equilateral  triangles.        (§  69.) 
And  since  the  angles  of  the  faces  are  all  equal,  the  trie- 
drals  whose  vertices  are  A,  B,  C,  and  D  are  equal. 

(§  464,  I.) 
Therefore,  ABCD  is  a  regular  tetraedron. 


304 


SOLID  GEOMETRY.  —  BOOK   VII. 


H 


2.  To  construct  a  regular  hexaedron,  or  cube 

Let  AB  be  the  given  edge. 

Construct  the  square  ABCD. 

Draw  AE,  BF,  CG,  and  DH,  each  equal     ^ 
to  AB  and  perpendicular  to  ABCD.  b)-. 

.  Dxaw  EF,  FG,  GR,  and  HE. 

Then,  AG  is  ?i  regular  hexaedron. 

For  by  construction,  its  faces  are  equal  squares. 

Hence,  its  triedrals  are  all  equal.       _  (§  464,  I.) 

3.  To  construct  a  regular  octaedron. 
Let  AB  be  the  given  edge. 
Construct  the  square  ABCD;  through 

its  centre  0  draw  EOF  perpendicular 
to  ABCD,  making  OE  —  OF  =  OA. 

Join  the  points  E  and  F  to  A,  B,  C, 
and  D. 

Then,  AEFC  is  a  regular  octaedron. 

For  draw  OA,  OB,  and  OD. 

Then  in  the  right  triangles  AOB,  AOE,  and  AOF, 

OA  =  OB  =  OE  =  OF. 
Therefore,     A  AOB  =  A  AOE  =  A  AOF. 
Whence,  AB  =  AE  =  AF. 

Then  the  eight  edges   terminating  at  E  and 
equal. 
Thus 


(§  63.) 

(§  QQ.) 

F  are  all 

(§  407,  L) 

the  twelve  edges  of  the  octaedron  are  all  equal, 

and  the  faces  are  equal  equilateral  triangles.  (§  69.) 

Again,  by  construction,  the  diagonals  of  the  quadrilateral 

BEDF  are  equal,  and  bisect  each  other  at  right  angles. 

Hence,  BEDF  is  a  square  equal  to  ABCD,  and  OA  is 

perpendicular  to  its  plane.  (§  400.) 

Therefore,  the "  pyramids  A-BEDF  and    E-ABCD   are 

equal ;  and  hence  the  polyedrals  A-BEDF  and  E-ABCD 

are  equal. 

In  like  manner,  any  two  polyedrals  are  equal. 
Hence,  AEFC  is  a  regular  octaedron. 


REGULAR   POLYEDRONS. 
4.    To  construct  a  regular  dodecaedron. 
d' 


305 


Let  AB  be  the  given  edge. 

Construct  the  regular  pentagon  ABODE  (Fig.  1). 

To  ABODE  join  five  equal  regular  pentagons,  so  in- 
clined as  to  form  equal  triedrals  at  the  vertices  A,  B,  0, 
D,  and  E.  (§  464,  I.) 

Then  there  is  formed  a  convex  surface  AK,  composed 
of  six  regular  pentagons,  as  shown  in  the  lower  portion  of 

Fig.  1. 

Construct  a  second  surface  A'K^  equal  to  AK,  as  shown 
in  the  upper  portion  of  Fig.  1. 

The  surfaces  AK  and  A'K^  may  be  combined  as  shown 
in  Fig.  2,  so  as  to  form  at  i^  a  triedral  equal  to  that  at  A, 
having  for  its  faces  the  regular  pentagons  about  the  ver- 
tices F  and  F'  in  Fig.  1.  (§  464,  I,) 

Then,  AK  is  a  regular  dodecaedron. 

For  since  G'  falls  at  G,  and  the  diedral  FG  and  the  face 
angles  FGH  and  FGU  (Fig.  2)  are  equal  respectively  to  the 
diedral  and  face  angles  of  the  triedral  F,  the  faces  about 
the  vertex  G  will  form  a  triedral  equal  to  that  at  F. 

Continuing  in  this  way,  it  may  be  proved  that  at  each  of 
the  vertices  H,  K,  etc.,  there  is  formed  a  triedral  equal  to 
that  at  F. 

Therefore,  AK  is  a  regular  dodecaedron. 


306                    SOLID   GEOMETRY. —BOOK   VII. 
5.    To  construct  a  regular  icosaedron. 
E^ ^F 


Fig.  1 


Fig.  3. 


Let  AB  be  the  given  edge. 

Construct  the  regular  pentagon  ABODE  (Fig.  1). 

At  its  centre  0  draw  OF  perpendicular  to  ABODE,  mak- 
ing AF  =  AB. 

Draw  AF,  BE,  OF,  DF,  and  EF. 

Then  F- ABODE  is  a  regular  pyramid  whose  lateral 
faces  are  equal  equilateral  triangles.  (§  69.) 

Construct  two  other  regular  pyramids,  A-BFEGH  and 
E-AFDKG,  each  equal  to  F- ABODE. 

Place  them  as  shown  in  the  upper  portion  of  Fig.  2,  so 
that  the  faces  ABE  and  AEF  of  A-BFEGH,  and  the 
faces  AEF  and  DEE  of  E-AFDKG,  shall  coincide  with 
the  corresponding  faces  of  F-ABODE. 

Then  there  is  formed  a  convex  surface  GO,  composed  of 
ten  equilateral  triangles. 

Construct  a  second  surface  G^O'  equal  to  GO,  as  shown  in 
the  lower  portion  of  Fig.  2. 

Then  the  surfaces  GO  and  G'O^  may  be  combined  as 
shown  in  Fig.  3,  so  that  the  edges  GR  and  HB  shall  coin- 
cide with  G'H^  and  H'B^,  respectively. 

For  since  the  diedrals  Aff,  E^W,  and  F'H'  are  equal  to 
the  diedrals  of  the  polyedral  F,  the  faces  about  the  vertices 
H  and  //'  may  be  made  to  form  a  polyedral  at  H  equal  to 
that  at  F.  (§  459.) 


REGULAR  POLYEDRONS. 


307 


Then  since  the  diedrals  FB,  AB,  'HB,  and  F'B  (Fig.  3) 
are  equal  to  the  diedrals  of  the  polyedral  F,  the  faces  about 
the  vertex  B  will  form  a  polyedral  equal  to  that  at  F. 

Continuing  in  this  way,  it  may  be  shown  that  at  each  of 
the  vertices  C,  D,  etc.,  there  is  formed  a  polyedral  equal  to 
that  at  F. 

Therefore,  (rC  is  a  regular  icosaedron. 

548.  ScH.  To  construct  the  regular  polyedrons,  draw  the 
following  diagrams  accurately  on  cardboard. 

Cut  the  figures  out  entire,  and  on  the  interior  lines  cut 
the  cardboard  half  through. 

The  edges  may  then  be  brought  together  so  as  to  form 
the  respective  solids. 


Tetraedron. 


Hexaedron. 


OCTAEDRON. 


DODECAEDKON. 


Icosaedron. 


SOLID   GEOMETRY.  —  BOOK  VII. 


EXERCISES. 

73.  If  the  volume  of  a  pyramid  whose  altitude  is  7  in.  is  686 
cu.  in.,  what  is  the  volume  of  a  similar  pyramid  whose  altitude  is 
12  in.  ? 

74.  If  the  volume  of  a  prism  whose  altitude  is.  9  ft.  is  171  cu.  ft., 
what  is  the  altitude  of  a  similar  prism  whose  volume  is  50f  cu.  ft.  ? 

75.  Two  bins  of  similar  form  contain,  respectively,  375  and  648 
bushels  of  wheat.  If  the  first  bin  is  3  ft.  9  in.  long,  what  is  the 
length  of  the  second  ? 

76.  A  pyramid  whose  altitude  is  10  in.,  weighs  24  lb.  At  what 
distance  from  its  vertex  must  it  be  cut  by  a  plane  parallel  to  its  base 
§0  that  the  frustum  cut  off  may  weigh  12  lb.  ? 

77.  An  edge  of  a  polyedron  is  56,  and  the  homologous  edge  of  a 
similar  polyedron  is  21.  The  area  of  the  entire  surface  of  the  second 
polyedron  is  135,  and  its  volume  is  162.  Find  the  area  of  the  entire 
surface,  and  the  volume,  of  the  first  polyedron. 

78.  The  area  of  the  entire  surface  of  a  tetraedron  is  147,  and  its 
volume  is  686.  If  the  area  of  the  entire  surface  of  a  similar  tetrae- 
dron is  48,  what  is  its  volume  ? 

79.  The  area  of  the  entire  surface  of  a  tetraedron  is  75,  and  its 
volume  is  500.  If  the  volume  of  a  similar  tetraedron  is  32,  what  i» 
the  area  of  its  entire  surface  ? 

80.  The  homologous  edges  of  three  similar  tetraedrons  are  3,  4, 
and  5,  respectively.  Find  the  homologous  edge  of  a  similar  tetrae' 
dron  equivalent  to  their  sum. 

81.  State  and  prove  the  converse  of  Prop.  XXVI. 

82.  The  volume  of  a  regular  tetraedron  is  equal  to  the  cube  of  its 
edge  multiplied  by  ^f^- 

83.  The  volume  of  a  regular  octaedron  is  equal  to  the  cube  of  its 
edge  multiplied  by  ^V2. 

84.  The  volume  of  a  regular  tetraedron  is  18  V2.  Find  the  area 
of  its  entire  surface. 

85.  The  sum  of  the  perpendiculars  drawn  to  the  faces  from  any 
point  within  a  regular  tetraedron  is  equal  to  the  altitude  of  the 
tetraedron. 


I 


BOOK  YIII. 

THE   CYLINDER,    CONE,    AND   SPHERE. 


DEFINITIONS. 

549.  A  cylindrical  surface  is  a  su 
moving  straight  line,  which  con- 
stantly intersects  a  given  curve, 
and  in  all  of  its  positions  is  paral- 
lel to  a  given  straight  line,  not  in 
the  plane  of  the  curve. 

Thus,  if  the  line  AB  moves  so 
as  to  constantly  intersect  the  curve 
AD,  and  is  constantly  parallel  to  the  line  MN,  not 
plane  of  the  curve,  it  generates  a  cylindrical  surface. 


N 


M 


face 

generated  by  a 

y|C 

B 

K 

-■■' 

>/ 

A 

^/ 

/ 

in  the 


550.  The  moving  straight  line  is  called  the  generatrix, 
and  the  curve  the  directrix;^  any  position  of  the  generatrix, 
as  EF,  is  called  an  element  of  the  surface. 


551.  A  cylinder  is  a  solid  bounded  by  a 
cylindrical  surface,  and  two  parallel  planes. 

The  parallel  planes  are  called  the  bases  of 
the  cylinder,  and  the  cylindrical  surface  the 
lateral  surface. 

The  altitude  of  a  cylinder  is  the  perpen- 
dicular distance  between  the  planes  of  its  bases. 

Note.     We  shall  use  the  phrase  "  element  of  a  cylinder 
nify  an  element  of  its  lateral  surface. 


"  to  sig- 


552.    It  follows  from  the  definition  of  §  551  that 

The  elements  of  a  cylinder  are  equal  and  parallel.   (§  418.) 


310 


SOLID   GEOMETRY.  —  BOOK   VIII. 


553.  A  rigJit  cylinder  is  a  cylinder  whose  elements  are 
perpendicular  to  its  bases. 

A  circular  cylinder  is  a  cylinder  whose  base  is  a  circle. 
The  axis  of  a  circular  cylinder  is  a  straight  line  drawn 
through  the  centre  of  its  base  parallel  to  its  elements. 

554.  A  right  circular  cylinder  is  called  a  cylinder  of  revo- 
lution ;  for  it  may  be  generated  by  the  revolution 
of  a  rectangle  about  one  of  its  sides  as  an  axis. 


555.  Similar  cylinders  of  revolution  are  cyl- 
inders generated  by  the  revolution  of  similar 
rectangles  about  homologous  sides  as  axes. 


Proposition  I.     Theorem. 

556.    A  section   of  a  cylinder  made  by  a  plane  passing 
through  an  element  is  a  parallelogram. 


liet  ABCD  be  ^  section  of  the  cylinder  ^C,  made  by  a 
plane  passing  through  the  element  AB. 
To  prove  ABCD  a  parallelogram. 

Draw  CE  in  the  plane  ABCD  parallel  to  AB. 
Then  CE  is  an  element  of  the  lateral  surface.        (§  552.) 
Therefore,   CE  must  'be   the   intersection   of   the  plane 
ABCD  with  the  lateral  surface  of  the  cylinder. 

Hence,  CE  coincides  with  CD,  and  CD  is  parallel  to  AB. 
Again,  AD  is  parallel  to  BC.  (§  417.) 

Therefore,  ABCD  is  a  parallelogram. 


THE   CYLINDER.  311 

557.  Cor.   A  section  of  a  right  cylinder  made  hy  a  plane 
passing  through  an  element  is  a  rectangle. 

Proposition  II.     Theorem. 

558.  The  bases  of  a  cylinder  are  equal. 


Let  AB'  be  a  cylinder. 

To  prove  its  bases  AB  and  A'B'  equal. 

Let  E',  F',  and  G'  be  any  three  points  in  the  perimeter 
of  A'B',  and  draw  the  elements  EE',  FF',  and  GG'. 

Draw  EF,  FG,  GE,  E'F\  F'G',  and  G'E'. 

Now,  EE'  and  FF'  are  equal  and  parallel.  (§  552.) 

Therefore,  EE'F'F  is  a  parallelogram.  (§  109.) 

Hence,  E'F'  =  EF.  (§  104.) 

Similarly,     E'G'  =  EG,  and  F'G'  =  FG. 

Therefore,  A  E'F'G'  =  A  EFG.  (§  69.) 

Then  the  base  A'B'  may  be  superposed  upon  AB  so  that 
the  points  E',  F',  and  G'  shall  fal-1  at  E,  F,  and  G. 

But  E'  is  any  point  in  the  perimeter  of  A'B'. 

Hence,  every  point  in  the  perimeter  of  A'B'  will  fall  in 
the  perimeter  of  AB,  and  A'B'  is  equal  to  AB. 

559.  Cor.  I.  The  sections  of  a  cylinder  made  by  two 
pai'allel  planes  cutting  all  its  elements  are  equal. 

For  they  are  the  bases  of  a  cylinder. 

560.  Cor.  II.  A  section  of  a  cylinder  made  by  a  plane 
parallel  to  the  base  is  equal  to  the  base. 


312 


SOLID   GEOMETRY. —BOOK   VIII. 


THE   CONE. 
Definitions. 

561.  A  conical  surface  is  a  surface  generated  by  a  moving 
straight  line,  which,  constantly  inter- 
sects a  given  curve,  and  passes  through 
a  given  point  not  in  the  plane  of  the 
curve. 

Thus,  if  the  line  OA  moves  so  as  to 
constantly  intersect  the  curve  ABC,  and 
constantly  passes  through  the  point  O, 
not  in  the  plane  of  the  curve,  it  gener- 
ates a  conical  surface. 

562.  The  moving  straight  line  is  called  the  generatrix, 
and  the  curve  the  directrix. 

The  given  point  is  called  the  vertex ;  and  any  position  of 
the  generatrix,  as  OB,  is  called  an  element  of  the  surface. 

563.  If  the  generatrix  be  supposed  indefinite  in  length, 
it  will  generate  two  conical  surfaces,  0-A'B'  C  and  0-ABC. 

These  are  called  the  upper  and  lower  nappes,  respectively. 

564.  A  cone  is  a  solid  bounded  by  a  conical  surface,  and 
a  plane  cutting  all  its  elements. 

The  plane  is  called  the  base  of  the  cone,  and 
the  curved  surface  the  lateral  surface. 

The  altitude  of  a  cone  is  the  perpendicular 
distance  from  the  vertex  to  the  plane  of  the 
base. 

565.  A  circular  cone  is  a  cone  whose  base  is  a  circle. 
The  axis  of  a  circular  cone  is  a  straight  line  drawn  from 

the  vertex  to  the  centre  of  the  base. 

566.  A  right  circular  cone  is  a  circular  cone  whose  axis 
is  perpendicular  to  its  base. 


THE   CONE. 


313 


567.  A  right  circular  cone  is  called  a  cone  of  revolution, 
for  it  may  be  generated  by  the  revolution  of  a 

right  triangle  about  one  of  its  legs  as  an  axis. 

568.  Simila?'  cones  of  revolution  are  cones 
generated  by  the  revolution  of  similar  right 
triangles  about  homologous  legs  as  axes. 

569.  A  frustum  of  a  cone  is  that  portion 
of  a  cone  included  between  the  base  and  a 
plane  parallel  to  the  base. 

The  altitude  of  a  frustum  is  the  perpen- 
dicular distance  between  the  planes  of  its 
bases. 


Proposition  III.     Theorem. 

570.    A  section  of  a  cone  made  by  a  plane  passing  through 
the  vertex  is  a  triangle. 


Let  OCD  be  a  section  of  the  cone  OAB,  made  by  a  plane 
passing  through  the  vertex  0: 
To  prove  OCD  a  triangle. 

Draw  straight  lines  in  the  plane  OCD  from  0  to  the 
points  C  and  D. 

These  lines  are  elements  of  the  lateral  surface.      (§  562.) 

Then  they  must  be  the  lines  of  intersection  of  the  plane 
OCD  with  the  lateral  surface  of  the  cone. 

Therefore,  OC  and  OD  are  straight  lines,  and  OCD  is  a 
triangle. 


S14  SOLID  GEOMETRY. —BOOK  VIII. 


Pkopositiox  IV.     Theorem. 

571.    A  section  of  a  circular  cone  made  hy  a  j^lccne  parallel 
to  the  base  is  a  circle. 


Let   A'B'C  be   a  section  of  the  circular  cone   S-ABC, 
made  by  a  plane  parallel  to  the  base. 
To  prove  A'B'C  a  circle. 

Let  the  axis  OS  intersect  the  plane  A'B'C  at  0'. 

Let  A'  and  B'  be  any  two  points  in  the  perimeter  A'B  C. 

Let  the  planes  determined  by  these  points  and  OS  inter- 
sect the  base  in  the  radii  OA  and  OB,  and  the  lateral 
surface  in  the  elements  SA  and  ^.  (§  570.) 

Then,  O'A'  is  parallel  to  OA,  and  O'B'  to  OB.       (§  417.) 

Therefore,  the  triangles  A'O'S  and  B'O'S  are  similar  to 
the  triangles  A  OS  and  BOS.  (§  258.) 

.,,,  O'A'     sa      .  O'B'     sa 

Whence,  _==^,and  — =  — . 

Then,  -^  =  ^- 

'  OA         OB 

But,  OA  =  OB.  (§  143.) 

Whence,  O'A'  =  O'B'. 

Now  A'  and  B'  are  any  two  points  in  the  perimeter  A'B'C. 

Therefore,  the  section  A'B'C  is  a  circle. 

572.  Cor.  The  axis  of  a  circular  cone  2^(isses  through  the 
centre  of  every  section  parallel  to  the  base. 


THE   SPHERE.  315 

THE   SPHERE. 
Definitions. 

573.  A  sphere  is  a  solid  bounded  by  a  surface,  all  points 
of  which  are  equally  distant  from  a  point  within  called  the 
centre. 

574.  A  radius  of  a  sphere  is  a  straight  line  drawn  from 
the  centre  to  the  surface. 

A  diameter  is  a  straight  line  drawn  through  the  centre, 
having  its  extremities  in  the  surface. 

575.  It  follows  from  the  definition  of  §  574  that 
All  radii  of  a  sphere  are  equal. 

Also,  all  its  diameters  are  equal,  since  each  is  the  sum  of 
two  radii. 

576.  A  sphere  may  be  generated  by  the  revolution  of  a 
semicircle  about  its  diameter  as  an  axis. 

577.  Two  spheres  are  equal  when  their  radii  are  equal. 
For  they  can  evidently  be  applied  one  to  the  other  so 

that  their  surfaces  shall  coincide  throughout. 

578.  Conversely,  the  radii  of  equal  spheres  are  equal. 

579.  A  straight  line  or  plane  is  said  to  be  tangent  to  a 
sphere  when  it  has  but  one  point  in  common  with  the  sur- 
face of  the  sphere. 

The  common  point  is  called  the  point  of  contact,  or  point 
of  tangency. 

580.  A  polyedron  is  said  to  be  inscribed  in  a  sphere  when 
its  vertices  lie  in  the  surface  of  the  sphere  ;  in  this  case  the 
sphere  is  said  to  be  circumscribed  about  the  polyedron. 

A  polyedron  is  said  to  be  circumscribed  about  a  sphere 
when  its  faces  are  tangent  to  the  sphere ;  in  this  case  the 
sphere  is  said  to  be  inscribed  in  the  polyedron. 


316  SOLID   GEOMETRY.  —  BOOK   VIII. 

Proposition  V.     Theokem. 

581.    A  section  of  a  sphere  made  by  a  jdane  is  a  cii'cle. 

p 


Let  ABC  be  a  section  of  the  sphere  AFC  made  by  a 
plane. 

To  prove  ABC  a  circle. 

Let  0  be  the  centre  of  the  sphere. 

Draw  OCf  perpendicular  to  the  plane  ABC. 

Let  A  and  B  be  any  two  points  in  the  perimeter  of  ABC, 
and  draw  0^,  OB,  a  A,  and  aB. 

Now,  OA  =  OB,  (§  575.) 

Whence,  O'A  =  O'B.  (§  408.) 

But  A  and  B  are  an?/  two  points  in  the  perimeter  of 
ABC. 

Therefore,  ABC  is  a  circle. 

582.  Def.  a  great  circle  of  a  sphere  is  a  section  made 
by  a  plane  passing  through  the  centre;  f* 

as  ABC.  /^^^1^\ 

A  small  circle  is  a  section  made  by  a  / 1 \ 

plane  which  does  not  pass  through  the  -^k::;^;;^o3Z^^ 

centre.  \  ^      I          / 

The  diameter  perpendicular  to  a  circle  ^^-J__-^ 
of  a  sphere  is  called  the  axis  of  the  cir- 
cle, and  its  extremities  are  called  the  poles. 

583.  Cor.  I.  The  axis  of  a  circle  of  a  sphere  2)(^sses 
through  the  centre  of  the  circle. 


THE   SPHERE.  317 

584.  Cor.  II.  All  great  circles  of  a  sphere  are  equal. 
For  their  radii  are  radii  of  the  sphere. 

585.  Cor.  III.  Every  great  circle  bisects  the  sphere  and 
its  surface. 

For  if  the  parts  be  separated,  and  placed  so  that  their 
plane  surfaces  coincide,  the  spherical  surfaces  falling  on 
the  same  side  of  this  plane,  the  two  spherical  surfaces  will 
coincide  throughout;  for  all  points  of  either  are  equally 
distant  from  the  centre. 

586.  Cor.  IV.    Any  two  great  circles  bisect  each  other. 
For  the  intersection  of  their  planes  passes  through  the 

centre  of  the  sphere,  and  hence  is  a  diameter  of  each  circle. 

587.  CoR.  V.  An  arc  of  a  great  circle,  less  than  a  semi- 
circumference,  may  be  drawn  betiveen  any  two  points  on  the 
surface  of  a  sphere,  and  but  one. 

For  the  two  points,  together  with  the  centre  of  the  sphere, 
determine  a  plane  which  intersects  the  surface  of  the  sphere 
in  the  arc  required. 

Note.  If  the  points  lie  at  the  extremities  of  a  diameter  of  the 
sphere,  an  indefinitely  great  number  of  arcs  of  great  circles  may  be 
drawn  between  them  ;  for  an  indefinitely  great  number  of  planes 
can  be  drawn  through  the  diameter. 

588.  Def.  The  distance  between  two  points  on  the  sur- 
face of  a  sphere  is  the  arc  of  a  great  circle, 

less  than  a  semi-circumference,  drawn  be-  _b^ 

tween  them. 

Thus,  the  distance  between  the  points  C 
and  B  is  the  arc  CED,  and  not  CFD. 


589.  Cor.  VI.  An  arc  of  a.  circle  may  be 
drawn  through  any  three  points  on  the  sur- 
face of  a  sphere. 

For  the  three  points  determine  a  plane,  which  intersects 
the  surface  of  the  sphere  in  the  arc  required. 


318  SOLID   GEOMETRY.  —  BOOK  VIII. 


Proposition  VI.     Theorem. 

590.   All  points  in  the  circumference  of  a  circle  of  a  sphere 
are  equally  distant  from  each  of  its  poles. 


Let  P  and  P'  be  the  poles  of  the  circle  ABC  of  the  sphere 
APC. 

To  prove  that  all  points  in  the  circumference  ABC  are 
equally  distant  ,(§  588)  froiu,^^  and  also  from^^ 

Let  A  and  B  be  any  two  points  in  the  circumference  ABC, 
and  draw  the  arcs  of  great  circles  PA  and  PB. 

Draw  the  axis  PP',  intersecting  the  plane.^^C  at  0. 
Draw  OA,  OB,  PA,  and  PB. 

Now  0  is  the  centre  of  the  circle  ABC.  (§  583.) 

Whence,  OA  =  OB.  (§  143.) 

Therefore,  chord  PA  =  chord  PB.  (§  407,  L) 

Whence,  arc  PA  =  arc  PB.  (§  157.) 

But  A  and  B  are  a7iy  two  points  in  the  circumference 
ABC. 

Hence,  all  points  in  the  circumference  ABC  are  equally 
distant  from  P. 

In  like  manner,  we  may  prove  that  all  points  in  the  cir- 
cumference ABC  are  equally  distant  from  P'. 

591.  Def.  The  polar  distance  of  a  circle  of  a  sphere  is 
the  distance  (§  588)  from  the  nearer  of  its  poles  to  the  cir- 
cumference of  the  circle. 

Thus,  the  j^olar  distance  of  the  circle  ABC  is  the  arc  PA. 


THE   SPHERE. 


319 


592.  Cor,  The  jpolar  distance  of  a  great  circle  is  a 
quadrant. 

Let  FA  be  the  polar  distance  of  the  p 

great  circle  ABC. 

To  i^rove  FA  a  quadrant  (§  146). 

Let  0  be  the  centre  of  the  sphere, 
and  draw  OA  and  OF. 

Then,  Z  FOA  is  a  right  angle. 

(§  398.) 

Therefore,  the  arc  FA  is  a  quadrant.  (§  191.) 

f 

593.  ScH.  The  term  quadrant^  in  Spherical  Geometry, 
usually  signifies  a  quadrant  of  a  great  circle. 

Proposition  VII.     Theorem. 

594.  If  a  point  on  the  surface  of  a  sphere  lies  at  a  quad- 
ranfs  distance  from  each  of  two  points  in  the  arc  of  a  great 
circle,  it  is  the  pole  of  that  a,rc. 


Let  P  be  a  point  on  the  surface  of  the  sphere  AC,  Sit  a, 
quadrant's  distance  from  each  of  the  points  A  and  B. 
To  prove  F  the  pole  of  the  arc  of  a  great  circle  AB. 

Draw  the  radii  OA,  OB,  and  OF. 

Then  since  the  arcs  FA  and  FB  are  quadrants,  the  angles 
FOA  and  FOB  are  right  angles.  (§  191.) 

Then,  PO  is  perpendicular  to  the  plane  OJ^-        (§  44)0.) 
Therefore,  P  is  the  pole  of  the  arc  AB. 


320  SOLID   GEOMETRY.  —  BOOK   VIll. 

Note.  If  the  two  points  lie  at  the  extremities  of  a  diameter,  the 
theorem  is  not  necessarily  true  ;  for  P  is  the  pole  of  only  one  of 
the  great  circles  which  can  be  drawn  through  the  points  A  and  C. 

Proposition  VIII.     Theorem. 

595.  A  plane  iperpendicular  to  a  radius  of  a  sphere  at  its 
extremity  is  tanfjet^  to  the  sphere. 


Let  the  plane  MN  be  perpendicular  to  the  radius  OA  of 
the  sphere  AC  Sit  its  extremity  A. 
To  prove  MN  tangent  to  the  sphere. 

Let  B  be  any  point  of  MN  except  A,  and  draw  OB. 
Then,  OB  >  OA.  (§  402.) 

Whence,  B  lies  without  the  sphere. 

Then  every  point  of  MN  except  A  lies  without  the  sphere, 
and  MN  is  tangent  to  the  sphere.  (§  579.) 

596.  CoR.  (Converse  of  Prop.  VIII.)  A  planejangent  to 
a  sphere  is  perpendicular  to  the  radius  drawn  to  the  point  of 
contact. 

Let  the  plane  MN  be  tangent  to  the  sphere  A  C. 

To  prove  that  MN  is  perpendicular  to  the  radius  OA 
drawn  to  the  point  of  contact. 

If  MN  is  tangent  to  the  sphere  at  A,  every  point  of  MN 
except  A  lies  without  the  sphere. 

Then  OA  is  the  shortest  line  that  can  be  drawn  from 
0  to  MN 

Whence,  OA  is  perpendicular  to  MN.  (§  402.) 


I 


THE  SPHERE.  321 

Proposition  IX.     Theorem. 
597.    A  sjjhev^  v^ai/  be  circumscribed  about  any  tetraedron. 
A 


Let  ABCD  be  any  tetraedron. 

To  prove  that  a  sphere  may  be  circumscribed  about  it. 

Draw  EK  in  the  face  ACD,  and  FK  in  the  face  BCD, 
perpendicular  to  CD  at  its  middle  points. 

Let  E  and  F  be  the  centres  of  the  circumscribed  circles 
of  the  triangles  ACD  and  BCD  (§  222). 

Draw  EG  and  FH  perpendicular  to  the  planes  ACD  and 
BCD. 

ISTow  the  plane  determined  by  EK  and  FK  is  perpen- 
dicular to  CD.  (§  400.) 

Therefore,  this  plane  is  perpendicular  to  each  of  the 
planes  ACD  and  BCD.  (§  444.) 

Then  EG,  being  perpendicular  to  the  plane  ^(7Z>,  lies  in 
the  plane  determined  by  EK  and  FK.  (§  441.) 

In  like  manner,  FH  lies  in  this  plane. 

Therefore,  EG  and  FH  must  meet  at  some  point,  as  0. 

Since  0  is  in  the  perpendicular  EG,  it  is  equally  distant 
from  A,  C,  and  D.  (§  407,  I.) 

And  since  it  is  in  the  perpendicular  FH,  it  is  equally 
distant  from  B,  C,  and  D. 

Hence,  0  is  equally  distant  from  A,  B,  C,  and  D ;  and  the 
sphere  described  with  0  as  a  centre,  and  OA  as  a  radius, 
will  be  cirQuniSCribed  about  the  tetraedron. 


322  SOLID   GEOMETRY.  —  BOOK  VIII. 

598.'  Cor.  I.  But  one  sphere  can  be  circumscribed  about  a 
given  tetraedron. 

For  the  centre  of  any  circumscribed  sphere  must  lie  in 
the  perpendicular  drawn  to  each  face  at  the  centre  of  its 
circumscribed  circle.  (§  408.) 

599.  Cor.  II.  The  planes  'perpendicular  to  the  edges  of  a 
tetraedron  at  their  middle  points  meet  in  a  common  point. 


Proposition  X.     Theorem. 
600.    A  sphere  may  be  ifiscTJibed  in  any  tetraedron. 


Let  ABCD  be  any  tetraedron. 

To  prove  that  a  sphere  may  be  inscribed  in  it. 

Draw  the  planes  OBC,  OCD,  and  ODB,  bisecting  the 
diedrals  ABCD,  ACDB,  and  ADBC,  respectively. 

Then  since  0  is  in  the  plane  OBC,  it  is  equally  distant 
from  the  faces  ABC  and  BCD.  (§  446.) 

In  like  manner,  0  is  equally  distant  from  the  faces  ACD 
and  BCD,  and  from  the  faces  ABD  and  BCD. 

Hence,  0  is  equally  distant  from  the  four  faces  of  the 
tetraedron ;  and  the  sphere  described  with  0  as  a  centre, 
and  the  perpendicular  from  0  to  either  face  as  a  radius, 
will  be  inscribed  in  the  tetraedron. 

601.  Cor.  The  planes  bisecting  the  diedrals  of  a  tetraedron 
meet  in  a  common  point. 


THE   SPHERE. 


323 


602.  Definitions.  The  angle  between  two  intersecting 
curves  is  the  angle  included  between  tangents  to  the  curves 
at  their  common  point. 

A  spherical  angle  is  the  angle  between  two  intersecting 
arcs  of  great  circles. 

Proposition  XI.     Theorem. 

603.  A  spherical  angle  is  measjired  by  the  arc  of  a  great 
circle  described  with  its  vertex  as  a  pole,  included  between  its 
sides  produced  if  necessary. 


Let  ABC  and  AB'C  be  arcs  of  great  circles  on  the  sur- 
face of  the  sphere  AC  whose  centre  is  0. 

Draw  ^i^and  ^'tangent  to  ABC  and  AB'G. 

Then  DAI/Ai  the  angle  between  the  arcs  ABC  and 
AB'C.  (§  602.) 

Draw  the  diameter  AC \  also,  draw  OB  and  OB'  in  the 
planes  ABC  and  AB'C  perpendicular  to  AC,  and  let  their 
plane  intersect  the  surface  of  the  sphere  in  the  arc  BB'. 

To  prove  that  Z.  DAD'  is  measured  by  the  arc  BB'. 

Z.  DAD^  is  the  plane  angle  of  the  diedral  BACB'.  (§  170.) 
Whence,  Z  DAD^  =  Z  BOB'.  (§  429.) 

But  Z  BOB'  is  measured  by  the  arc  BB'.  (§  192.) 

Therefore,  Z  DAD'  is  measured  by  the  arc  BB'. 

604.  CoR.  The  angle  between  two  arcs  of  great  circles  is 
the  plane  angle  of  the  diedral  formed  by  their  planes. 


324    .  SOLID   GEOMETllY.  —  BOOK   VIII. 

SPHERICAL  POLYGONS  AND  SPHERICAL  PYRAMIDS. 

Definitions. 

605.  A  spherical  polygon  is  a  portion  of  the  surface  of  a 
sphere  bounded  by  three  or  more  arcs 

of  great  circles  \  as  ABCD. 

The  bounding  arcs  are  called  the  sides 
of  the  spherical  polygon. 

The  angles  of  the  spherical  polygon 
are  the  spherical  angles  (§  602)  formed 
by  the  adjacent  sides ;  and  their  vertices 
are  called  the  vertices  of  the  spherical 
polygon. 

A  diagonal  is  an  arc  of  a  great  circle  joining  any  two 
vertices  which  are  not  consecutive. 

606.  The  planes  of  the  sides  of  a  spherical  polygon  form 
a  polyedral,  0-ABCD,  whose  vertex  is  the  centre  of  the 
sphere,  and  whose  face  angles  AOB,  BOC,  etc.,  are  meas- 
ured by  the  sides  AB,  BC,  etc.,  of  the  spherical  polygon 
(§  192). 

A  spherical  polygon  is  called  convex  when  its  correspond- 
ing polyedral  is  convex  (§  456). 

607.  Since  the  sum  of  the  face  angles  of  any  convex 
polyedral  is  less  than  four  right  angles  (§  463),  the  sum  of 
their  measures  is  less  than  a  circumference. 

That  is,  the  sum  of  the  sides  of  a  convex  spherical  polygon 
is  less  than  the  circumference  of  a  great  circle. 

608.  A  spherical  pyramid  is  the  solid  bounded  by  a  spher- 
ical polygon  and  the  planes  of  its  sides  ;  as  0-ABCD  (§  605). 

The  centre  of  the  sphere  is  called  the  vertex  of  the  spher- 
ical pyramid,  and  the  spherical  polygon  is  called  its  base. 

609.  The  sides  of  a  spherical  polygon,  being  arcs,  are 
usually  measured  in  degrees. 


I 


THE   SPHERE. 


325 


610.  A  spherical  triangle  is  a  spherical  polygon  of  three 
sides. 

It  is  called  isosceles,  equilateral,  or  right-angled  in  the 
same  cases  as  a  plane  triangle. 

611.  If,  with  the  vertices  of  a  spherical  triangle  as  poles, 
arcs  of  great  circles  be  described,  a 

spherical  triangle  is  formed  which  is 
called  the  polar  triangle  of  the  first. 
Thus,  if  ^,  B,  and  C  are  the  poles 
of  the  arcs  B'C,  C'A',  and  A'B%  then 
A'B'C  is  the  polar  triangle  of  ABC. 

612.  The  circumferences  of  which 
B'C,  C'A',  and  A'B'  are  arcs,  form 

by  their  intersection  eight  spherical  triangles. 

Four  of  these,  i.e.,  A'B'C,  A'B'C",  A'B"C',  and  A'B"C", 
lie  on  the  hemisphere  represented  in  the  figure,  and  the 
others  lie  on  the  opposite  hemisphere. 

Of  these  eight  spherical  triangles,  that  is  the  polar 
triangle  in  which  the  vertex  A'  homologous  to  A,  lies  on 
the  same  side  of  BC  as  the  vertex  A ;  and  similarly  for  the 


613.  Two  spherical  polygons  are  equal  wlien  they  can  be 
applied  one  to  the  other  so  as  to  coincide  throughout. 

614.  Two  spherical  polygons  are  equal  when  the  sides 
and  angles  of  one  are  equal  respectively  to  the  homologous 
sides  and  angles  of  the  other,  if  the  equal  parts  are  arranged 
in  the  same  order. 

Thus,  the  *  spherical  triangles 
ABC  and  A'B'C  are  equal  if  the 
sides  AB,  BC,  and  CA  are  equal 
to  A'B',  B'C,  and  C'A%  respec- 
tively, and  the  angles  A,  B,  and  C  to  the  angles  A%  B', 
and  C ;  for  they  can  evidently  be  applied  one  to  the  other 
so  as  to  coincide  throughout. 


c 


326  SOLID   GEOMETRY.  —  BOOK    VIII. 

615.  Two  spherical  polygons  are  said  to  be  symmetrical 
when  the  sides  and  angles  of  one  are  equal  respectively  to 
the  homologous  sides  and  angles  of  ^^  ^« 
the  other,  if  the  equal  parts  are  ar- 
ranged in  the  reverse  order.                        /^^  ^' 

Thus,  the  spherical  triangles  ABC    ^  ^^^' 

and  A'B'C  are  symmetrical  if  the  sides  AB,  BC,  and  CA 
are  equal  to  A'B',  B'C,  and  C'A%  respectively,  and  the 
angles  A^  B,  and  C  to  the  smgles  A',  B',  and  C. 

616.  It  is  evident  that  in  general  two  symmetrical 
spherical  triangles  cannot  be  applied  one  to  the  other  so 
as  to  coincide  throughout.     (Compare  §  636.) 

Proposition  XII.     Theorem. 

617.  If  one  spherical  tria'^Qle  is  the  polar  triangle  of  an- 
other, then  the  second  spherical  triangle  is  the  polar  triangle 
of  the  first. 


Let.^^,^a^  be  the  polar  triangle  of  ABC. 

To  prove  that  ABC  is  the  polar  triangle  of  A'B'C. 

Now  B  is  the  pole  of  the  arc  A'C.  .  (§  611.) 

Whence,  A'Aiq^  at  a  quadrant's  distance  from  B.  (§  592.) 
Again,  C  is  the  pol^  of  the  arc  A'B'. 
Whence,  A'  lies  at  a  quadrant's  distance  from  C. 
Therefore,  A'  is  the  pole  of  the  arc  BC.  (§  594.) 

In  like  manner,  it  may  be  proved  that  B'  is  the  pole  of 
the  arc  CA,  and  C  of  the  arc  AB. 

Then,  ABC  is  the  polar  triangle  of  A'B'C^^,-^^ 


THE  SPHERE.  327 

For  the  homolc/gous  vertices  A  and  A^  lie  on  the  same  side 
of  B'C  (§  012),  and  similarly  for  the  remaining  vertices. 

618.  Def.  Two  spherical  triangles,  each  of  which  is  the 
polar  triangle  of  the  other,  are  called  pola?'  triangles. 

Proposition  XIII.     Theorem. 

619.  Tn  two  polar  triangles,  each  angle  of  one  is  meas- 
ured by  the  supplement  of  the  side  lying  opposite  the  homolo- 
gous angle  of  the  other. 

X 


Let  a,  h,  c,  and  a',  V,  c'Menote  the  ^es,  expressed  in 
degrees,  and  A,  B,  C,  and  A^/Bi^^'  the  angles,  also  ex- 
pressed in  degrees,  of  the  polar  triangles  ABC  and  A'B'C. 
To  prove 

A  =  180°  -  a',     B  =  180°  -h\     C  =  180°  -  c% 
A'  =  180°  —  a,      B'=  180°  -  b,      C  ^  180°  -  c. 

Produce  the  arcs  AB  and  AC  to  meet  the  arc  B'C  at  D 
and^. 

Then  since  B'  is  the  pole  of  the  arc  AE,  and  C  of  the 
arc  AD,  the  arcs  B'E  and  CD  are  quadrants.  (§  592.) 

Therefore,       arc  B'E  +  arc  CD  =  180°. 

That  is,.  arc  DE  -f  arc  B'C  =  180°. 

But  A  is  the  pole  of  the  arc  B'C. 

Whence,  the  angle  A  is  measured  by  the  arc  DE.  (§  603.) 

Therefore,  A -\- a'  =  180°. 

Whence,  A  =  180°  —  a\      ■ 

In  like  manner,  the  theorem  may  be  proved  for  any  angle 
of  either  triangle. 


328  SOLID  GEOMETRY.  — BOOK  VIII. 


Proposition   XIV.     Theorem. 

620.    The  sum  of  any  two  sides  of  a  spherical  triangle  is 
greater  than  the  third  side. 


Let  ABC  be  a  spherical  triangle  on  the  surface  of  a  sphere 
whose  centre  is  0. 

To  prove  AB  ^  AC>  BC. 

Draw  OAj  OB,  and  OC;  then  in  the  triedral  0-ABC, 

Z  AOB  -{- Z  AOC>  ZBOa  (§  462.) 

But  the  sides  AB,  AC,  and  BC  are  the  measures  of  the 
angles  AOB,  AOC,  and  BOC,  respectively.  (§  192.) 

Therefore,  AB  -{- AC>  BC. 

'  In  like  manner,  we  may  prove 

AB  -^  BC>  AC,  Sind  AC  -\-  BC>  AB. 

Proposition  XV.     Theorem. 

621.    The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  two,  and  less  than  six,  right  angles. 


Let  ABC  be  a  spherical  triangle. 


THE   SPHERE.  329 

To  prove     A -\- B  -\-  C>  180°,  and  <  540^. 

Let  A'B'C  be  the  polar  triangle  of  ABC,  and  denote  its 
sides  by  a',  h',  and  c'. 

Then,  A  =  180°  -  a% 

B  =  180°  -  h', 
and  C  =  180°  -  c\  (§  619.) 

Adding  these  equations,  we  have 

^4_j5  +  C  =  540°- (^'+Z»'+c').       .        (1) 
Whence,  A-^B-^C  <  540°. 

Again,  the  sum  of  the  sides  of  the  spherical  triangle 
A'B'C  is  less  than  the  circumference  of  a  great  circle. 

(§  607.) 
That  is,  a'  -\-b'  -{-c'  <  360°. 

Whence  by(l),    A-j-B-{-C>  180°. 

622.  CoR.  I.   A  spherical  triangle  may  have  one,  two,  or 
three  right  angles,  or  one,  two,  or  three  obtuse  angles. 

623.  Def.   a  spherical  triangle  having  two  right  angles 
is  called  a  bi-rectangular  triangle. 

A  spherical  triangle  having  three  right  angles  is  called  a 
tri-rectangular  triangle. 

• 

624.  CoR.  II.    If  three  plaribs  he  passed  through  the  centre 

of  a  sphere  in  such  a  way  that  each  is 

perpendicular  to  the  other  tivo,  the  sur-  ^-^y^^^^-^ 

face  is  divided  into  eight  equal  tri-rec-  /^  f  \  \    \. 

tangular  triangles.                          •  L-'T ^' ''~^^-\ 

For  each  angle  of  either  spherical  tri-  V^~^"'  T   -j — J 

angle  is  a  right  angle.  \1\  ''   <^ y 

Also,  each  side  of  either  triangle  is  a  ^^^^ 
quadrant.                                         (§  191.) 

Hence,  the  spherical  triangles  are  all  equal.  (§  614.) 

""     625.    CoR.  III.    The  surface  of  a  sphere  is  eight  times  the 
surface  of  a  tri-rectangular  triangle. 


330  SOLID   GEOMETRY.  —  BOOK   YIII. 

626.  Def.  Two  spherical  polygons  on  the  same  sphere, 
or  equal  spheres,  are  said  to  be  mutually  equilateral,  or 
mutually  equiangular,  when  the  sides  or  angles  of  one  are 
equal  respectively  to  the  homologous  sides  or  angles  of  the 
other,  whether  taken  in  the  same  or  in  the  reverse  order. 


Proposition  XVI.     Theorem. 

627.  If  two  spherical  triangles  on  the  same  sphere,  or 
equal  spheres,  are  mutually  equiangular,  their  polar  tri- 
angles are  mutually  equilateral. 


Let  ABC  and  DEF  be  mutually  equiangular  spherical 
triangles  on  the  same  sphere,  or  on  equal  spheres  ;  the 
angles  A  and  D  being  homologous. 

To  prove  that  their  polar  triangles,  A'B'C  and  VE'F', 
are  mutually  equilateral. 

The  angles  A  and  D  are  measured  by  the  supplements 
of  the  sides  B'C  and  E'F',  respectively.  (§  619.) 

But  by  hypothesis,       A  A=A  B. 

Whence,  B'C  =  E'F'.  (§  33,  2.) 

In  like  manner,  any  two  homologous  sides  of  A  B'C  and 
I/E'F'  may  be  proved  .equal. 

Therefore,  A' B'C  and  I/E'E'  are  mutually  equilateral. 

628.  Cor.  (Converse  of  Prop.  XVI.)  If  two  spherical  tri- 
angles on  the  same  sphere,  or  equal  spheres,  are  m^utually 
equilateral,  their  p>olar  triangles  are  mutually  equiangular. 

(The  proof  is  left  to  the  student ;  compare  §  627.) 


THE   SPHERE.  331 

Proposition  XVII.     Theorem. 

629.    Jf  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  have  two  sides  and  the  included  angle  of  one  equal 
respectively  to  two  sides  and  the  included  angle  of  the  other, 
I.    They  are  equal  if  the  equal  parts  occur  in  the  same 
order. 

II.    They  are  symmetrical  if  the  equal  j)Cbrts  occur  in  the 
reverse  order. 

D' 


I.  Let  ABC  and  DSF  be  spherical  triangles  on  the  same 
sphere,  or  equal  spheres,  having 

AB  =  DE,  AC  =  DF,  and  Z  A  =  Z  D. 
To  prove  ABC  and  DEF  equal. 

Superpose  ABC  upon  DFF  so  that  Z  A  shall  coincide 
with  Z  D ;  the  side  AB  falling  upon  DE,  and  AC  upon  DF. 

Then,  since  AB  =  DE  and  AC  =  DF,  B  will  fall  at  E, 
and  C  at  i^;  and  the  side  BC  will  coincide  with  EF.  (§  587.) 

Hence,  ABC  and  DEF  coincide  throughout,  and  are  equal. 

II.  Let  ABC  and  D'E'F'  be  spherical  triangles  on  the 
same  sphere,  or  equal  spheres,  having 

AB  =  D^E',  AC  =  D^F',  smd  ZA=ZD'. 
To  prove  ABC  and  D'E'F'  symmetrical. 

Construct  the   spherical   triangle  DEF  symmetrical   to 
D'E'F',  having  DE  =  D'E',  DF  =  D'F',  and  ZD=ZD'. 
Then  in  the  spherical  triangles  ABC  and  DEF,  we  have 

AB  =  DE,  AC  =  DF,  and  Z  A  =ZD. 
Whence,  ABC  and  DEF  are  equal.  (§  629,  I.) 

Therefore,  ABC  is  symmetrical  to  D'E'F'. 


332  SOLID   GEOMETRY.  —  BOOK  YIII. 

Proposition  XVIII.     Theokem. 

630.  If  two  S2)he7'ical  triangles  on  the  same  sphere,  or 
equal  spheres,  have  a  side  and  two  adjacent  angles  of  one 
equal  respectively  to  a  side  and  two  adjacent  angles  of  the 
other, 

I.  They  are  equal  if  the  equal  parts  occur  in  the  same 
order. 

II.  Tliey  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 

(The  proof  is  left  to  the  student ;  compare  §  629.) 

Proposition  XIX.     Theorem. 

631.  If  two  spherical  triangles  on  the  same  sphere,  or 
equal  spheres,  are  mutually  equilateral,  they  are  m^utually 
equiangular. 


Let  ABC  and  DBF  be  mutually  equilateral  spherical 
triangles  on  equal  spheres,  the  sides  BC  and  EF  being 
homologous. 

To  prove  ABC  and  DEF  mutually  equiangular. 

Let  0  and  0'  be  the  centres  of  the  respective  spheres. 

Draw  OA,  OB,  OC,  O'D,  O'E,  and  O'F. 

The  triedrals  0-ABC  and  C-DEF  have  their  homolo- 
gous face  angles  equal.  (§  192.) 

Therefore,         diedral  OA  =  diedral  CD.  (§  465.) 

But  the  spherical  angles  BAC  and  EDF  are  the  plane 
angles  of  the  diedrals  OA  and  O'D.  (§  604.) 

Whence,  Z  BAC  =  Z.  EDF.  (§  432.) 


THE   SPHERE.  333 

In  like  manner,  any  two  homologous  angles  of  ABC  and 
DEF  may  be  proved  equal. 

Whence,  ABC  and  DEF  are  mutually  equiangular. 

632.  CoR.  If  two  spherical  triangles  on  the  same  sphere, 
or  equal  splieresj  are  mutually  equilateral, 

1.  They  are  equal  if  the  equal  parts  occur  in  the  same  order. 

2.  They  are  symmetrical  if  the  equal  ptarts  occur  in  the 
reverse  order. 

Proposition  XX.     Theorem. 

633.  If  two  spherical  triangles  on  the  same  sphere,  or  equal 
spheres,  are  mutually  equiangular,  they  are  mutually  equilat- 
eral. 

A'  D' 


Let  ABC  and  DEF  be  mutually  equiangular  spherical 
triangles  on  the  same  sphere,  or  equal  spheres. 
To  prove  ABC  and  DEF  mutually  equilateral. 

Let  A'B'C  be  the  polar  triangle  of  ABC,  and  D^E'F'  of 
DEF. 

Then  since  ABC  and   DEF  are  mutually  equiangular, 
A'B'C  and  D'E'F'  are  mutually  equilateral.  (§  627.) 

Then  A!B' C  and  D^E'E'  are  mutually  equiangular.  (§  631.) 
Therefore,  their  polar  triangles,  ABC  and  DEF,  are  mutu- 
ally equilateral.  (§  627.) 

634.    Cor.  If  two  spherical  triangles  on  the  same  sphere, 
or  equal  spheres,  aA'e  mutually  equiangular, 

1.  They  are  equal  if  the  equal  parts  occur  in  the  same  order. 

2.  They  are  symmetrical  if  the  equal  parts  occur  in  the 
reverse  order. 


su 


SOLID   GEOMETRY.— BOOK   VIII. 


Proposition  XXI.     Theorem. 

635.    In  an  isosceles  spiertccd  triangle,  the  angles  oppo- 
site the  equal  sides  are  eqn^l.      ' 


In  the  spherical  triangle  ABC,  let  AB  =  AC. 
To  prove  Z  B  =  Z  C. 

Let  AD  be  an  arc  of  a  great  circle  bisecting  the  side  BC. 
/Then  in  the  spherical  triangles  ABD  and  ACI),  the  side 
AD  is  common  ;  also,  AB  =  AC,  and  BD  =  CD. 

Then  ABD  and  ACD  are  mntually  equiangular.   (§  631.) 


Whence, 


ZB  =  ZC. 


636.  Cor.  I.  Two  sijmmetrlcal  (§  615)  isosceles  spherical 
triangles  are  equal ;  for  they  can  be  applied  one  to  the 
other  so  as  to  coincide  throughout. 

637.  Cor.  II.  (Converse  of  Prop.  XXI.)  If  two  angles 
of  a  spherical  triangle  are  equal,  the  sides  opposite  are  equal. 

In  the  spherical  triangle  ABC,  let 
ZB  =  ZC. 
To  prove  AB  =  AC. 

Let  AB'C   be  the  polar   triangle  of 

ABC. 

Then,  AB'  is  the  supplement  of  Z  C,     ^ 

and  AC  of  Z  B.  (§  619.) 

Whence,  A'B'  =  A'C  (§  33,  2.) 

Therefore,  Z  C  =  Z  B'.  (§  635.) 

But  AB  is  the  supplement  of  Z  C ,  and  AC  oi  Z  B'. 
Whence,  AB  =  AC. 


THE  SPHERE.  335 


Proposition  XXII.     Theorem. 

638.    In  any  spherical  triangle,  the  greater  side  lies  oppo- 
site the  greater  angle. 


In  the  spherical  triangle  ABC,  let  A  ABC  be  >  AC. 
To  prove  AC>  AB. 

Let  BD  be  an  arc  of  a  great  circle  making  Z  CBD  =  AC. 
Then,  '  BD  =  CD.  (§  637.) 

But,  AD  +  BD  >  AB.  (§  620.) 

Therefore,  AD  -\-  CD  >  AB. 

That  is,  AC>  AB. 

639.    Cor.    (Converse  of  Prop.  XXII.)    In  any  spherical 
triangle,  the  greater  angle  lies  opposite  the  greater  side. 
In  the  spherical  triangle  ABC,  let  ^C  be  >  AB. 
To  prove  AABC>AC 

li  AABC  were  <  A  C,  AC  would  be  <  AB.         (§  638.) 

And  if  A  ABC  were  equal  to  A  C,  AC  would  be  equal  to 
AB.  (§  637.) 

But  each  of  these  conclusions  is  contrary  to  the  hypothesis 
that  ^C  is  >  AB. 

Hence,  AABC>AC. 

EXERCISES. 

1.  If  the  sides  of  a  spherical  triangle  are  77°,  123°,  and  95°,  how 
many  degrees  are  there  in  each  angle  of  its  polar  triangle  ? 

2.  If  the  angles  of  a  spherical  triangle  are  86°,  131°,  and  68°,  how 
many  degrees  are  there  in  each  side  of  its  polar  triangle  ? 


336  SOLID   GEOMETRY.  —  BOOK   VIII. 


Proposition  XXIII.     Theorem. 

640.  The  shortest  line  on  the  surface  of  a  sphere  between 
two  given  points  is  the  arc  of  a  great  circle,  not  greater 
than  a  semi-circumference,  which  joins  the  points. 


Let  AB  be  an  arc  of  a  great  circle,  not  greater  than  a 
semi-circumference,  which  joins  the  given  points  A  and  B. 

To  prove  AB  the  shortest  line  on  the  surface  of  the 
sphere  between  A  and  B. 

Let  C  be  any  point  in  the  arc  AB. 

Let  DCF  and  ECG  be  arcs  of  small  circles,  having  A 
and  B  respectively  as  poles,  and  AC  and  ^C  as  polar 
distances. 

The  arcs  DCF  aiid  ECG  have  only  the  point  C  common. 

For  let  F  be  any  other  point  in  tlie  arc  DCF,  and  draw 
the  arcs  of  great  circles  AF  and  BF. 

Then,  AF  -{-  BF  >  AC  -^  BC.  (§  620.) 

Subtracting  arc  AF  from  the  first  member  of  the  ine- 
quality, and  its  equal  AC  from  the  second  member,  we  have 
BF  >  BC. 

Therefore,  F  lies  without  the  small  circle  ECG,  and  the 
arcs  DCF  and  ECG  have  only  the  point  C  common. 

We  will  next  prove  that  the  shortest  line  on  the  surface 
of  the  sphere  from  ^  to  ^  must  pass  through  C. 

Let  ADEB  be  any  line  drawn  on  the  surface  of  the 
sphere  between  A  and  B,  not  passing  through  C,  and  cut- 
ting the  arcs  DCF  and  ECG  at  D  and  E,  respectively. 

Then,  whatever  the  nature  of  the  line  AD,  it  is  evident 
that  an  equal  line  can  be  drawn  from  A  to  C. 


THE   SPHERE.  337 

In  like  manner,  whatever  the  nature  of  the  line  BE,  an 
equal  line  can  be  drawn  from  B  to  G. 

Hence,  a  line  can  be  drawn  from  ^  to  ^  passing  through 
C,  equal  to  the  sum  of  the  lines  AD  and  BE,  and  conse- 
quently less  than  the  line  ADEB  by  the  portion  DE. 

Therefore,  no  line  which  does  not  pass  through  C  can 
be  the  shortest  line  between  A  and  B. 

But  C  is  any  point  in  the  arc  AB. 

Hence  the  shortest  line  from  A  to  B  must  pass  through 
every  point  of  AB. 

That  is,  the  arc  of  a  great  circle  AB  is  the  shortest  line 
which  can  be  drawn  on  the  surface  of  the  sphere  between 
A  and  B. 

EXERCISES. 

3.  Any  point  in  the  arc  of  a  great  circle  bisecting  a  spherical 
angle  is  equally  distant  (§  588)  from  the  sides  of  the  angle. 

[Prove,  by  §§  G29,  II.,  and  33,  1,  the  equality  of  the  arcs  of  great 
circles  joining  the  point  to  the  poles  of  the  sides  of  the  angle.] 

4.  State  and  prove  the  converse  of  Ex.  3. 

5.  The  sum  of  the  angles  of  a  spherical  hexagon  is  greater  than 
8,  and  less  than  12,  right  angles. 

6.  The  sum  of  the  angles  of  a  spherical  polygon  of  n  sides  is 
greater  than  2  n  —  4,  and  less  than  2  n,  right  angles. 

7.  The  sides  opposite  the  equal  angles  of  a  hi-rectangular  tri- 
angle are  quadrants. 

8.  State  and  prove  the  converse  of  Ex.  7. 

9.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of  the 
remaining  sides. 

10.  The  arc  of  a  great  circle  drawn  from  the  vertex  of  an  isos- 
celes spherical  triangle  to  the  middle  point  of  the  base,  is  perpen- 
dicular to  the  base,  and  bisects  the  vertical  angle. 

11.  How  many  degrees  are  there  in  the  polar  distance  of  a  circle, 
whose  plane  is  5  V2  units  from  the  centre  of  the  sphere,  the  diame- 
ter of  the  sphere  being  20  units  ? 

12.  The  polar  distance  of  a  circle  of  a  sphere  is  60°.  If  the 
diameter  of  the  circle  is  6,  find  the  diameter  of  the  sphere,  and 
the  distance  of  the  circle  from  its  centre. 


338 


SOLID   GEOMETRY.— BOOK  YIII. 


MEASUREMENT   OF  SPHERICAL  POLYGONS. 

641.  Def.    a  liine  is  a  portion  of  the  surface  of  a  sphere 
bounded    by   two    semi-circumferences    of 
PTeat_circ1es  ;  as  ACBD. 

The  angle  of  the  lune  is  the  angle  in- 
cluded between  its  bounding  arcs. 

642.  It  is  evident  that  two  lunes  on  the 
same  sphere,  or  equal  spheres,  are  equal 
when  their  angles  are  equal. 

643.  A  spherical  wedge  is  the  solid  bounded  by  a  lune 
and  the  planes  of  its  bounding  arcs. 

The  lune  is  called  the  base  of  the  spherical  wedge. 


Proposition  XXIV.     Theorem. 

644.    The  spherical  triantjles  corresponding  to  a  pair  of 
vertical  triedrals  are  symmetrical. 


JjQt  AOA',  BOB',  and  COC  be  diameters  of  the  sphere  ^C. 
To  prove  the  spherical  triangles  ABC  and  AB'C  sym- 
metrical. 

The  angles  AOB,  BOC,  and  CO  A  are  equal  respectively 
to  the  angles  A' OB',  B'OC,  and  C'OA\  (§  39.) 

Then,  AB  =  A'B',  BC  =  B'C,  and  CA  =  C'A'.    (§  192.) 

But  the  equal  parts  of  ABC  and  A'B'C  are  arranged  in 
the  reverse  order. 

Therefore,  ABC  and  A'B'C  are  symmetrical.    (§  Q>32,  2.) 


THE  SPHERE.  339 

Proposition  XXV.     Theorem. 
645.    Tivo  syinmetrlcal  spherical  triangles  are  equivalent. 


Let  AA',  BB',  and  CC  be  diameters  of  the  sphere  AB. 
Then  the  spherical  triangles  ABC  and  A'B'C  are  sym- 
metrical. (§  644.) 
To  prove           area  ABC  =  area  A'B'C. 

Let  P  be  the  pole  of  the  small  circle  passing  through  the 
poihts  A,  B,  and  C,  and  draw  the  arcs  of  great  circles  FA^ 
PB,  and  PC. 

Then,  PA  =  PB  =  PC.  (§  590.) 

Draw  the  diameter  of  the  sphere  PP\ 

Also,  draw  the  arcs  of  great  circles  P'A',  P'B',  and  P'C. 

Then  the  spherical  triangles  PAB  and  P'A'B'  are  sym- 
metrical. (§  644.) 

But  the  spherical  triangle  PAB  is  isosceles. 

Therefore,  PAB  is  equal  to  P'A'B'.  (§  636.) 

In  like  manner,  we  may  prove  PBC  equal  to  P'B'C, 
and  PCA  equal  to  P'C  A'. 

Then  the  sum  of  the  areas  of  the  triangles  PAB,  PBC, 
and  PCA  is  equal  to  the  sum  of  the  areas  of  P'A'B',  P'B'C\ 
and  P'C  A'. 

That  is,  area  ABC  =  area  A'B'C 

646.  ScH.  If  P  and  P'  fall  without  the  spherical  tri- 
angles ABC  and  A'B'C,  we  should  take  the  sum  of  the 
areas  of  two  isosceles  spherical  triangles,  diminished  b^ 
the  area  of  a  third. 


340 


SOLID   GEOMETKY.  —  BOOK   VIII. 


Proposition  XXVI.     Theorem. 

647.    Tivo  lunes  on  the  same  sphere,  or  equal  spheres,  are 
to  each  other  as  their  angles. 

Note.     The  word  "  ^wne,"  in  the  ahove  statement,  signifies  the 
area  of  the  lune. 

Case  I.     When  the  angles  are  commensurable. 


Let  ACBD  and  AC  BE  be  two  lunes,  whose  angles  CAD 
and  CAE  are  commensurable. 

ACBD       /.CAD 


To  prove 


ACBE       Z  CAE 


Let  CAa  be  a  common  measure  of.  the  angles  CAD  and 
CAE,  and  let  it  be  contained  5  times  in  CAD,  and  3  times 
in  CAE. 

A  CAD  ^  5 

Z  CAE      3' 


Then, 


(1) 


Producing  the  several  arcs  of  division  of  the  angle  CAD 

to  B,  the  lune  ACBD  will  be  divided  into  5  parts,  and  the 

lune  ACBE  into  3  parts,  all  of  which  parts  will  be  equal. 

(§  642.) 
ACBD      5 

(2) 


Then, 


ACBE      3 

From  (1)  and  (2),  we  have, 

ACBD  ^  Z  CAD 
ACBE       Z  CAE' 


THE   SPHERE.  341 

Case  IT.    When  the  arigles  are  incommensurable. 


Let  ACBD  and  AC  BE  be  two  liiiies,  whose  angles  CAD 
2^n(i'CAE  are  incommensurable. 

ACBD       Z  CAD 


To  prove 


AC  BE       Z  CAE 


Let  Z  CAD  be  divided  into  any  number  of  equal  parts, 
and  let  one  of  these  parts  be  applied  to  Z  CAE  as  a 
measure. 

Since  CAD  and  CAE  are  incommensurable,  a  certain 
number  of  thh  parts  will  extend  from  AC  to  AE',  leaving 
a  remainder  E'AE  less  than  one  of  ^the  p^rts. 

Produce  the  arc  AE'  to  B. 

Now  let  the  number  of  subdivisions  of  the  angle  CAD  be 
indefinitely  increased. 

Then  the  magnitude  of  each  part  will  be  indefinitely 
diminished,  and  the  remainder  E'AE  will  approach  the 
limit  0. 

Then,      ^ — -  will  approach  the  limit  , 

'      ACBE'  ^^  ACBE' 

-,  Z  CAD      ...  ,    , ,     V    -4-  ^  G^D 

Z  CAE'  approach  the  limit  ^  ^^^ . 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 
ACBD       Z  CAD 


Whence 


'  ACBE       Z  CAE' 


342  SOLID   GEOMETRY.  —  BOOK   VIII. 

648.  Cob.  I.  The  surface  of  a  lune  is  to  the  surface  of  • 
the  sphere  as  the  angle  of  the  lune  is  to  four  right  angles. 

For  the  surface  of  a  sphere  may  be  regarded  as  a  lune 
whose  angle  is  equal  to  four  right  angles. 

649.  CoK.  II.  Let  L  denote  the  area  jof  jk  lune ;  A  the 
numerical  measure  of  its  angle  referred  to  a  right  angle  as 
the  unit ;  and  T  the  area  of  a  tri-rectangular  triangle. 

Then  the  area  of  the  surface  of  the  sphere  is  8  T.  (§  625.) 

Whence,  A^  =  A ,  (§648.) 

>  8T        4  y^         J 

That  is,  L=2AyX:^. 

Hence,  if  the  unit  of  'measure  for  angles  is  the  right  angle, 
the  area  of  a  lune  is  equal  to  twice  its  angle,  viultiplied  hy 
the  area  of  a  trv-rectangular  triangle. 

Thus,  if  the  area  of  the  surface  of  a  sphere  is  72,  the 
area  of  a  tri-rectangular  triangle  is  \  of  72,  or  9. 

Then,  if  the  angle  of  a  lune  on  this  sphere  is  50°,  or  ^  of 
a  right  angle,  its  area  is  J/  of  9,  or  10. 

650.  ScH.    It  may  be  proved,  exactly  as  in  §  647,  that 
The  volume  of  a  spherical  wedge  is  to  the  volume  of  the 

sphere  as  the  angle  of  the  lune  which  forms  its  base  is  to  four 
right  angles. 

It  follows  from  the  above  that 

If  the  unit  of  measure  for  angles  is  the  right  angle,  the 
volume  of  a  spherical  wedge  is  equal  to  twice  the  angle  of 
the  lune  which  forms  its  base,  multiplied  by  the  volume  of 
a  tri-rectangular  pyramid. 

Note.  A  tri-rectangular  pyramid  is  a  spherical  pyramid  whose 
base  is  a  tri-rectangular  triangle. 

651.  Def.  The  spherical  excess  of  a  spherical  triangle  is 
the  excess  of  the  sum  of  its  angles  above  two  right  angles. 

Thus,  if  the  angles  of  a  spherical  triangle  are  65°,  80°,  and 
95°,  its  spherical  excess  is  m°  +  80°  +  95°  -  180°,  or  60°. 


THE  SPHERE.  343 


Proposition  XXVII.     Theorem. 

652.  If  the  unit  of  measure  for  angles  is  the  right  angle, 
the  area  of  a  spherical  triangle  is  equal  to  its  spherical  excess, 
multiplied  by  the  area  of  a  tri-rectangidar  triangle. 


Let  A,  B,  and  C  denote  the  numerical  measures  of  the  an- 
gles of  the  spherical  triangle  ABC  referred  to  a  right  angle 
as  the  unit ;  and  T  the  area  of  a  tri-rectangular  triangle. 

To  prove  area  ABC  =  {A  +  B  -^  C —2)x  T. 

Complete    the    circumferences    ABA'B',    ACA'C,    and 
BCB'C;  and  draw  the  diameters  AA',  BB',  and  CC. 
Then  since  ABA'C  is  a  lune  whose  angle  is  A,  we  have 

area  ABC  -\-  area  A'BC  =  2  A  x  T  (%  649).         ( 1 ) 
And  since  BAB'C  is  a  lune  whose  angle  is  B, 

area  ABC  +  area  AB'C  =  2B  xT.  (2) 

Kow  A'B'C  and  ABC  are  symmetrical.  (§  644.) 

Whence,  area  A'B'C  =  area  ABC.  (§  645.) 

Adding  area  ABC  to  both  members,  we  have 

area  ABC  +  area  A'B'C  =  area  of  lune  CBC'A 

=  2CxT.  (3) 

•  Adding  (1)-,  (2),  and  (3),  and  observing  that  the  sum  of 
the  areas  of  ABC,  A'BC,  AB'C,  and  A'B'C  is  equal  to  the 
area  of  the  surface  of  a  hemisphere,  or  4  T,  we  have 
2  area  ABC  -\-4:T=(2A-\-2B  +  2C)xT. 
Then,     area  ABC  -{- 2  T  =  (A -\-  B  -\-  C)x  T. 
Or,  area  ABC  =  (A -\-  B -\-  C  —  2)x  T. 


344  SOLID   GEOMETRY.  —  BOOK   VIII. 

653.  ScH.  I.  Let  it  be  required  to  find  the  area  of  a 
spherical  triangle  whose  angles  are  105°,  80°,  and  95°,  on 
a  sphere  the  area  of  whose  surface  is  144  sq.  in. 

The  spherical  excess  of  the  spherical  triangle  is  100°,  or 
J^o-  referred  to  a  right  angle  as  the  unit. 

And  the  area  of  a  tri-rectangular  triangle  is  ^  of  144,  or 
18  sq.  in. 

Hence,  the  required  area  is  \^  of  18,  or  20  sq.  in. 

654.  ScH.  II.   It  may  be  proved,  as  in  §  652,  that 

If  the  unit  of  measure  for  angles  is  the  right  angle,  the 
volume  of  a  triangular  spherical  pyramid  is  equal  to  the 
spherical  excess  of  its  base,  'multiplied  by  the  volu7ne  of  a 
tri-rectangular  pyramid. 

EXERCISES. 

13.   Find  the  area  of  a  spherical  triangle  whose  angles  are  103°, 
112°,  and  127°,  on  a  sphere  the  area  of  whose  surface  is  160. 
"^14.   Find    the  volume    of    a    triangular  spherical   pyramid   the 
angles  of  whose  base  are  92°,  119°,  and  134°;  the  volume  of  the 
sphere  being  192. 

15.  What  is  the  volume  of  a  spherical  wedge  the  angle  of  whose 
base  is  127°  30',  if  the  vol.ume  of  the  sphere  is  112  ? 

16.  The  area  of  a  lune  is  28f  sq.  in.  If  the  area  of  the  surface 
of  the  sphere  is  120  sq.  in.,  what  is  the  angle  of  the  lune  ? 

17.  What  is  the  ratio  of  the  areas  of  two  spherical  triangles  on 
the  same  sphere,  whose  angles  are  94°,  135°,  and  146°,  and  87°,  105°, 
and  118°,  respectively  ? 

18.  The  area  of  a  spherical  triangle  two  of  whose  angles  are  78° 
and  99°,  is  34^.  If  the  area  of  the  surface  of  the  sphere  is  234,  what 
is  the  other  angle  ? 

19.  The  volume  of  a  triangular  spherical  pyramid  the  angles 
of  whose  base  are  105°,  126°,  and  147°,  is  60^;  what  is  the  volume 
of  the  sphere  ? 

20.  If  two  straight  lines  are  tangent  to  a  sphere  at  the  same  point, 
their  plane  is  tangent  to  the  sphere. 

21.  The  sum  of  the  arcs  of  great  circles  drawn  from  any  point 
within  a  spherical  triangle  to  the  extremities  of  any  side,  is  less  than 
the  sum  of  the. other  two  sides  of  the  triangle. 


3 


THE  SPHERE.  345 

Proposition  XXVIII.     Theorem. 

655.  Jf  the  unit  of  measure  for  angles  is  the  right  angle, 
the  area  of  any  spherical  polygon  is  equal  to  the  sum  of  its 
angles,  diminished  hy  as  many  times  two  right  angles  as  the 
figure  has  sides  less  two,  multiplied  hy  the  area  of  a  tri-rect- 
angular  triangle. 


Let  K  denote  the  area  of  any  spherical  polygon  ;  n  the 
number  of  its  sides ;  s  the  sum  of  its  angles  referred  to 
a  right  angle  as  the  unit ;  and  T  the  area  of  a  tri-rect- 
angular  triangle. 

To  prove  K  =  [s  —  2  (n  —  2)^  X  T.' 

The  spherical  polygon  may  be  divided  into  spherical 
triangles  by  drawing  diagonals  from  any  vertex  ;  the  num- 
ber of  such  spherical  triangles  being  equal  to  the  number  of 
sides  of  the  spherical  polygon,  less  two. 

Now  the  area  of  each  spherical  triangle  is  equal  to  the 
sum  of  its  angles,  less  two  right  angles,  multiplied  by  T. 

(§  652.) 

Hence,  the  sum  of  the  areas  of  the  spherical  triangles  is 
equal  to  the  sum  of  their  angles,  diminished  by  as  many 
times  two  right  angles  as  there  are  triangles,  multiplied 
byT. 

But  the  number  of  triangles  is  ?^  —  2. 

Therefore,  the  area  of  the  spherical  polygon  is  equal  to 
the  sum  of  its  angles,  diminished  hy  n  —  2  times  two  right 
angles,  multiplied  by  T. 

That  is,  K=ls-2{n-  2)]  X  T. 


346  SOLID   GEOMETRY.  —  BOOK  VIII. 

656.  ScH.   It  may  be  proved,  as  in  §  655,  that 

If  the  unit  of  measure  for  angles  is  the  right  angle,  the 
volume  of  any  spherical  pyramid  is  equal  to  the  sum  of  the 
angles  of  its  base,  di7ninished  hy  as  many  times  two  right 
angles  as  the  base  has  sides  less  two,  multiptlied  by  the  volume 
of  a  tri-rectangular  pyramid. 

657.  Cor.  Let  P  denote  the  volume  of  a  spherical  pyra- 
mid ;  and  let  K  denote  the  area  of  the  base,  n  the  number 
of  its  sides,  and  s  the  sum  of  its  angles  referred  to  a  right 
angle  as  the  unit. 

Let  V  represent  the  volume  of  the  sphere ;  S  the  area  of 
its  surface  ;  T  the  area  of  a  tri-rectangular  triangle ;  and  T' 
the  volume  of  a  tri-rectangular  pyramid. 

Then,  P  =  ls-2{n-2)-]X  T',  (§  656.) 

and  K={s  —  2{n  —  2)]  X  T.  (§  655.) 

Also,  r  =  8  T',  and  .S  =  8  T. 

Whence  by  division, 

•p  _  T^     .v_^r_ 

K       T'^^^    ST' 

P  V 

Therefore,  £_  =  Jl_. 

-ST        S 

That  is,  the  volume  of  a  spherical  pyramid  is  to  its  base 
as  the  volume  of  the  sphere  is  to  its  surface. 

EXERCISES. 

22.  Find  the  area  of  a  spherical  hexagon  whose  angles  are  120°, 
139°,  148°,  155°,  162°,  and  167°,  on  a  sphere  the  area  of  whose  sur- 
face is  280. 

23.  Find  the  volume  of  a  pentagonal  spherical  pyramid  the 
angles  of  whose  base  are  109°,  128°,  137°,  153°,  and  158°;  the  volume 
of  the  sphere  being  180. 

24.  The  arcs  of  great  circles  bisecting  the  angles  of  a  spherical 
triangle  meet  in  a  point  equally  distant  from  the  sides  of  the 
triangle.     (Exs.  3,  4,  p.  337.) 

25.  A  circle  may  be  inscribed  in  any  spherical  triangle. 


THE  sriIERE.  347 

26.  State  and  prove  the  theorem  for  spherical  triangles  analogous 
to  Prop.  IX.,  I.,  Book  I. 

27.  State  and  prove  the  theorem  for  spherical  triangles  analogous 
to  Prop,  v.,  Book  I. 

28.  State  and  prove  the  theorem  for  spherical  triangles  analogous 
to  Prop.  LI.,  Book  I. 

29.  The  volume  of  a  quadrangular  spherical  pyramid,  the  angles 
of  whose  base  are  110°,  122°,  135°,  and  146°,  is  12i  cu.  ft.  What  is 
the  volume  of  the  sphere  ? 

30.  The  area  of  a  spherical  pentagon,  four  of  whose  angles  are 
112°,  131°,  138°,  and  168°,  is  27.  If  the  area  of  tlie  surface  of  the 
sphere  is  120,  what  is  the  other  angle  ? 

31.  If  the  side  AB  oi  a.  spherical  triangle  ABC  is  equal  to  a 
quadrant,  and  the  side  BC  is  less  than  a  quadrant,  prove  that  Z  A  is 
less  than  90°. 

32.  If  PJ.,  PB,  and  PC  are  three  equal  arcs  of  great  circles 
drawn  from  a  point  P  to  the  circumference  of  a  great  circle  ABCj 
prove  that  P  is  the  pole  of  ABC. 

33.  The  spherical  polygons  corresponding  to  a  pair  of  vertical 
polyedrals  are  symmetrical. 

34.  Either  angle  of  a  spherical  triangle  is  greater  than  the  dif- 
ference between  180°  and  the  sum  of  the  other  two  angles. 

35.  If  a  polyedron  be  circumscribed  about  each  of  two  equal 
spheres,  the  volumes  of  the  polyedrons  are  to  each  other  as  the  areas 
of  their  surfaces. 

36.  If  ABC  and  A'B'C  are  a  pair  of  polar  triangles  on  a  sphere 
whose  centre  is  O,  prove  that  the  radius  OA'  is  perpendicular  to  the 
plane  OBC, 

37.  The  intersection  of  two  spheres  is  a  circle,  whose  centre  lies 
in  the  line  joining  the  centres  of  the  spheres,  and  whose  plane  is 
perpendicular  to  this  line. 

38.  The  distance  between  the  centres  of  two  spheres,  whose  radii 
are  25  in.  and  17  in.,  respectively,  is  28  in.  Find  the  diameter  of 
their  circle  of  intersection,  and  the  distance  of  its  plane  from  the 
centre  of  each  sphere. 


BOOK   IX. 


MEASUREMENT    OP     THE     CYLINDER, 
AND    SPHERE. 


CONE, 


THE   CYLmDER. 


Definitions. 


658.  A  prism  is  said  to  be  inscribed  in  a  cylinder  when 
its  bases  are  inscribed  in  the  bases  of  the  cylinder. 

A  prism  is  said  to  be  circumscribed  about  a  cylinder  when 
its  bases  are  circumscribed  about  the  bases  of  the  cylinder. 

The  lateral  area  of  a  cylinder  is  the  area  of  its  lateral 
surface. 

A  right  section  of  a  cylinder  is  a  section  perpendicular  to 
its  elements. 

659.  If  a  regular  polygon  be  inscribed  in,  or  circumscribed 
about,  a  circle,  and  the  number  of 

its  sides  be  indefiuitely  increased, 
its  perimeter  and  area  approach  the 
circumference  and  area  of  the  circle 
respectively  as  limits  (§  363). 

Hence,  if  a  prism  whose  base  is  a 
regular  polygon  be  inscribed  in,  or 
circumscribed  about  a  circular  cylin- 
der (§  553),  and  the  number  of  its 
faces  be  indefinitely  increased, 

1.  The  lateral  area  of  the  prism 
approaches  the  lateral  area  of  the  cylinder  as  a  limit. 

2.  The  volume  of  the  prism  approaches  the  volume  of  the 
cylinder  as  a  limit. 

348 


MEASUREMENT   OF  THE   CYLINDER.  349 

3.  The  perimeter  of  a  right  section  of  the  prism  ap- 
proaches the  perhneter  of  a  right  section  of  the  cylinder  as 
a  limit. 

Proposition  I.     Theorem. 

660.  The  lateral  area  of  a  circular  cylinder  is  equal  to 
'  the  perimeter  of  a  right  section,  multiplied  by  an  element. 


Let  S  denote  the  lateral  area,  P  the  perimeter  of  a  right 
section,  and  ^  an  element,  of  a  circular  cylinder. 
To  prove  S  =  PxE. 

Inscribe  in  the  cylinder  a  prism  whose  base  is  a  regular 
polygon. 

Let  S'  denote  its  lateral  area,  and  P'  the  perimeter  of  a 
right  section. 

Then  since  the  lateral  edge  of  the  prism  is  E,  we  have 

S'  =  F'xE.  (§485.) 

Now  let  the  number  of  faces  of  the  prism  be  indefinitely 
increased. 

Then,  S'  approaches  the  limit  S, 

and  P'  xE  approaches  the  limit  P  xE. 

(§  659,  1,  3.) 

By  the  Theorem  of  Limits,  these  limits  are  equal.  (§  188.) 

Therefore,  S  =  PxE. 

661.    CoR.  I.   The  lateral  area  of  a  cylinder  of  revolution 
is  equal  to  the  perimeter  of  its  base  multiplied  by  its  altitude. 


350  SOLID   GEOMETRY.  —  BOOK  IX. 

662.  Cor.  II.  Let  S  denote  the  lateral  area,  T  the  total 
area,  J£  the  altitude,  and  H  the  radius  of  the  base,  of  a 
cylinder  of  revolution. 

Then,  S^^ttRH.  (§368.) 

And  T  =  2  ttRH  +  2  irB^  (§  371.) 

=  2  7rE(II-\-E). 

Proposition  II.     Theorem. 

663.  The  volume  of  a  cirmdar  cylinder  is  equal  to  the 
product  of  its  base  and  altitude. 


Let  V  denote  the  volume,  B  the  area  of  the  base,  and  JI 
the  altitude,  of  a  circular  cylinder. 
To  prove  V  =  BxH. 

Inscribe  in  the  cylinder  a  prism  whose  base  is  a  regular 
polygon ;  let  V  denote  its  volume,  and  B'  the  area  of  its  base. 
Then  since  the  altitude  of  the  prism  is  H,  we  have 

V  =  B'  X  H.  (§  507.) 

Now  let  the  number  of  faces  of  the  prism  be  indefinitely 
increased. 

Then,  V  approaches  the  limit  V.  (§  659, 2.) 

And  B'  XH  approaches  the  limit  B  X  H.  {^  363, 11.) 

Therefore,  V  =  B  X  H.  (§  188.) 

664.    CoR.    Let  V  denote  the  volume,  H  the  altitude,  and 
B,  the  radius  of  the  base,  of  a  circular  cylinder. 

Then,  V  =  7rEUI.  (§371.) 


MEASUREMENT   OF   THE   CYLINDER. 


351 


pROPOsiTiox  III.     Theorem. 

665.  TJie  lateral  or  total  areas  of  two  similar  cylinders  of 
revolution  are  to  each  other  as  the  squares  of  their  altitudes, 
or  as  the  squares  of  the  radii  of  their  bases  ;  and  their  vol- 
umes are  to  each  other  as  the  cubes  of  their  altitudesj  or  as 
the  cubes  of  the  radii  of  their  bases. 


Let  S  and  s  denote  the  lateral  areas,  T  and  t  the  total 
areas,  V  and  v  the  volumes,  H  and  h  the  altitudes,  and  H 
and  r  the  radii  of  the  bases,  of  two  similar  cylinders  of 
revolution  (§  555). 

S       T      H^      E^ 


To  prove 


and 


s         t         h^         r''' 

V        A*        r^ ' 
The  generating  rectangles  are  similar. 
H      E 


Whence, 


Then, 
S  _2  irEH 
s         2  irrh 


h        r 
H  -\-E 

h-\-r 


(§  253,  11.) 
(§  239.) 


(§  662)  = 


n 


X  -  =  —  =  — 


t  2  Trr  (h  +  7-)     ^^^  ''^        r         r        r^        h^ 


and 


rrEm 

7r7-h 


(§  664)  =^x^  =  ^  =  ^. 
^       r^       r        r^        h^ 


352  SOLID   GEOMETKY.  —  BOOK  IX. 

THE   CONE. 
Definitions. 

666.  A  pyramid  is  said  to  be  inscribed  in  a  cone  when 
its  base  is  inscribed  in  the  base  of  the  cone,  and  its  vertex 
coincides  with  the  vertex  of  the  cone. 

A  pyramid  is  said  to  be  circumscribed  about  a  cone  when 
its  base  is  circumscribed  about  the  base  of  the  cone,  and  its 
vertex  coincides  with  the  vertex  of  the  cone. 

A  frustum  of  a  pyramid  is  said  to  be  hiscribed  in  a 
frustum  of  a  cone  when  its  bases  are  inscribed  in  the 
bases  of  the  frustum  of  the  cone. 

A  frustum  of  a  pyramid  is  said  to  be  circumscribed  about 
a  frustum  of  a  cone  when  its  bases  are  circumscribed  about 
the  bases  of  the  frustum  of  the  cone. 

The  lateral  area  of  a  cone,  or  frustum  of  a  cone,  is  the 
area  of  its  lateral  surface. 

The  slant  height  of  a  cone  of  revolution  is  the  straight 
line  drawn  from  the  vertex  to  any  point  in  the  circumfer- 
ence of  the  base. 

The  slant  height  of  a  frustum  of  a  cone  of  revolution  is 
that  portion  of  the  slant  height  of  the  cone  included  between 
the  bases  of  the  frustum. 

667.  It  may  be  proved,  exactly  as  in  .  Jk 

§  659,  that  /fn\    ' 

If  a  pyramid  whose  base  is  a  regular  /a/  /  /  jl\ 

polygon  be  inscribed  in,  or  circumscribed  y^^;,'/i^£f  ~--k  \ 
about,  a  circular  cone  (§  565),  and  the  \/  /  /  ]'Y\ 
number  of  its  faces  be  indefinitely  in-  ^^^^j-s^^^^''^''''^ 
creased, 

1.  The  lateral  area  of  the  pyramid  approaches  the  lat- 
eral area  of  the  cone  as  a  limit. 

2.  The  volume  of  the  pyramid  approaches  the  volume  of 
the  cone  as  a  limit. 


MEASUREMENT  OF  THE   CONE.  353 

668.  It  follows  from  §  6G7  that 

If  a  frustum  of  a  jtijramid  whose  base  is  a  regular  polygon 
be  inscribed  in,  or  circumscvihed  about,  a  frustum  of  a  circu- 
lar cone,  and  the  number  of  its  faces  be  ijidefinitelij  increased, 

1.  The  lateral  area  of  the  frustum  of  the  pyramid  ap- 
proaches the  lateral  area  of  the  frustum,  of  the  cone  as  a  limit. 

2.  The  volume  of  the  frustum  of  the  -pyramid  approaches 
the  volume  of  the  frustum  of  the  cone  as  a  limit. 

Proposition  IV.     Theorem. 

669.  The  lateral  area  of  a  cone  of  revolution  is  equal  to 
the  circumference  of  its  base,  multiplied  by  one-half  its  slant 
height. 


Let  S  denote  the  lateral  area,  C  the  circumference  of  the 
base,  and  L  the  slant  height,  of  a  cone  of  revolution. 
To  prove  S=Cx\L. 

Circumscribe    about  the  cone  a  regular  pyramid;   let  S 
denote  its  lateral  area,  and  C  the  perimeter  of  its  base. 

Then  since  the  sides  of  the  base  of  the  pyramid  are  bi- 
sected at  their  points  of  contact,  the  slant  height  of  the  pyr- 
amid is  the  same  as  the  slant  height  of  the  cone.       (§  515.) 
Therefore,  S'  =C'  X^L.  (§  523.) 

Now  let  the  number  of  faces  of  the  pyramid  be  indefi- 
nitely increased. 

Then,  S'  approaches  the  limit  S.  (§  667,  1.) 

And      C  X^L  approaches  the  limit  C  X  ^  L.  (§  363,  I.) 
Therefore,  S^Cx^L.  (§  188.) 


364  SOLID  GEOMETRY.  —  BOOK  IX. 

670.  Cor.  Let  S  denote  the  lateral  area,  T  the  total 
area,  L  the  slant  height,  and  li  the  radius  of  the  base,  of  a 
cone  of  revolution. 

Then,  S  =  2^-1^  X  i  L  (%  368)  =  ttEL. 

And    T  =  TT  ML  +  TT  ^2  (§  371)  =  tt  B(L  +  B). 

Pkoposition  v.     Theorem. 

671.  The  volume  of  a  circular  cone  is  equal  to  one-third 
the  product  of  its  base  and  altitude. 


Let  V  denote  the  volume,  B  the  area  of  the  base,  and  H 
the  altitude,  of  a  circular  cone. 

To  prove  V  =  ^  B  X  H. 

Inscribe  in  the  cone  a  pyramid  whose  base  is  a  regular 
polygon;  let  V^  denote  its  volume,  and  B'  the  area  of  its 
base. 

Then,  V  =  ^B'  X IL  .  (§  527.) 

Now  let  the  number  of  faces  of  the  pyramid  be  indefi- 
nitely increased. 

Then,  F'  approaches  the  limit  V.  (§  667,  2.) 

And      ^B'  X  H  approaches  the  limit  ^  B  X  H. 

(§363,n.) 
Therefore,  V  =  ^  B  X  JT.  (§  188.) 

672.  CoR.  Let  V  denote  the  volume,  //  the  altitude,  and 
B  the  radius  of  the  base,  of  a  circular  cone. 

Then,  V=lTrBm,  (§371.) 


MEASUREMENT   OF   THE   CONE.  355 


Proposition  YI.     Theorem. 

673.  The  lateral  or  total  areas  of  two  similar  cones  of 
Involution  are  to  each  other  as  the  squares  of  their  slant 
Jt eights,  or  as  the  squares  of  their  altitudes,  or  as  the  squares 
of  the  radii  of  their  bases  ;  and  their  volumes  are  to  each 
other  as  the  cubes  of  their  slant  heights,  or  as  the  cubes  of 
their  altitudes,  or  as  the  cubes  of  the  radii  of  their  bases. 


Let  S  and  s  denote  the  lateral  areas,  T  and  t  the  total 
areas,  V  and  v  the  volumes,  L  and  I  the  slant  heights,  H 
and  h  the  altitudes,  and  R  and  r  the  radii  of  the  bases,  of 
two  similar  cones  of  revolution  (§  568). 

S  ^  T  ^TJ  ^  m  ^W- 
s        t         I'         h' 

.  V     //     ir^     B^ 


To  prove  ^         ^         ,.^ 


The  generating  triangles  are  similar. 


I         r         h  I  -{-  r 


Whence, 
Then, 

~s ~ ~^:m~        ^^ ^^  ^ -'v^v~'^~i^~~¥' 

T~     ^r(l  +  r)     ^^^^^^-T^T~'^~'P~    h^' 
and 


356  SOLID   GEOMETRY.  —  BOOK  IX. 


Proposition^  VII.     Theorem. 

674.  The  lateral  area  of  a  frustum  of  a  cone  of  revolution 
is  equal  to  one-half  the  sum  of  the  circumferences  of  its  bases, 
multiplied  by  its  slant  height. 


Let  S  denote  the  lateral  area,  C  and  c  the  circumferences 
of  the  bases,  and  L  the  slant  height,  of  a  frustum  of  a  cone 
of  revolution. 

To  prove  S  =  \{C  -^  c)  X  L. 

Circumscribe  about  the  frustum  a  frustum  of  a  regular 
pyramid. 

Let  S'  denote  its  lateral  area,  and  C  and  c'  the  perime- 
ters of  its  bases. 

Then  since  the  sides  of  the  bases  of  the  frustum  of  the 
pyramid  are  bisected  at  their  points  of  contact,  the  slant 
height  of  the  frustum  of  the  pyramid  is  the  same  as  the 
slant  height  of  the  frustum  of  the  cone.  (§  515.) 

Therefore,  S\  =  1{C'  +  c')x  L.  (§524.) 

Now  let  the  number  of  faces  of  the  frustum  of  the  pyra- 
mid be  indefinitely  increased. 

Then,  S'  approaches  the  limit  S.  (§  668,  1.) 

And 
^{C  -\-  c')xL  approaches  the  limit  \  {C  -\-  c)  X  L. 

(§  363,  I.) 

Therefore,  S^=  \{C  -\-c)  xL.  (§  188.) 


MEASUREMENT   OF   THE    CONE.  357 

675.  Cor.  I.  The  lateral  area  of  a  frustum  of  a  cone  of 
revolution  is  equal  to  the  circMmference  of  a  section  equally 
distant  fro  ni  its  bases,  multiplied  by  its  slant  height. 

676.  Cor.  II.  Let  S  denote  the  lateral  area,  L  the  slant 
height,  and  R  and  r  the  radii  of  the  bases,  of  a  frustum 
of  a  cone  of  revolution. 

Then,  ^  =  1  {2'kB  +  ^irr)  X  X  (§  368)  =  7r(B  +  r)  L. 

Proposition  VIII.     Theorem. 

677.  The  volume  of  a  frustum  of  a  circular  cone  is  equal 
to  the  sum  of  its  bases  and  a  mean  proportional  betweeri  its 
bases,  multiplied  by  one-third  its  altitude. 


Let  V  denote  the  volume,  B  and  b  the  areas  of  the  bases, 
and  If  the  altitude,  of  a  frustum  of  a  circular  cone. 
To  prove       V  =  (B  -{- b -^  -Vlf^b)  X^H. 

Inscribe  in  the  frustum  a  frustum  of  a  pyramid  whose 
base  is  a  regular  polygon ;  let  V  denote  its  volume,  and  B' 
and  b'  the  areas  of  its  bases. 


Then,         F'  =  {B'  +b'  -^  ^ B'  Xb')X^  H.  (§  532.) 

Now  let  the  number  of  faces  of  the  frustum  of  the  pyra- 
mid be  indefinitely  increased. 

Then,  V  approaches  the  limit  V.  (§  668,  2.) 

And  {B'  -^b'  -\-  VB'  X  b')  X^H  approaches  the  limit 

{B-\-b  +  V^Xb)  X^H.  (§  363,  11.) 

Whence,      V={B-\-b^  VB^b)  X  I H.  (§  188.) 


358  SOLID   GEOMETRY.  —  BOOK   IX. 

678.  CoK.  Let  V  denote  the  volume,  H  the  altitude,  and 
R  and  7*  the  radii  of  the  bases,  of  a  frustum  of  a  circular 
cone. 

Then,  B  =  irR^  and  b  =  irr^  (§  371.) 

Whence,  Vi?  X  h  =  V^r^i^V^  =  irBr. 

Therefore,    V  =  (ttE^  +  vrr^  +  'jtEt)  x  -^  // 
=  1 TT  (i^2  +  r^  +  Br)  H. 

EXERCISES. 

» 

1.  Find  the  lateral  area,  total  area,  and  volume  of  a  cylinder  of 

revolution,  the  diameter  of  whose  base  is  18,  and  whose  altitude  is  16. 

2.  Find  the  lateral  area,  total  area,  and  volume  of  a  cone  of 
revolution,  the  radius  of  whose  base  is  7,  and  whose  slant  height 
is  25. 

3.  Find  the  lateral  area,  total  area,  and  volume  of  a  frustum  of 
a  cone  of  revolution,  the  diameters  of  whose  bases  are  16  and  6,  and 
whose  altitude  is  12. 

4.  Find  the  altitude  and  diameter  of  the  base  of  a  cylinder  of 
revolution,  whose  lateral  area  is  168  n  and  volume  504  n. 

5.  Find  the  volume  of  a  cone  of  revolution,  whose  slant  height 
is  29  and  lateral  area  580  ti. 

6.  Find  the  lateral  area  of  a  cone  of  revolution,  whose  volume  is 
320  TT  and  altitude  15. 

7.  Find  the  volume  of  a  cylinder  of  revolution,  whose  total  area 
is  170  Tt  and  altitude  12. 

8.  The  altitude  of  a  cone  of  revolution  is  27,  and  the  radius  of 
its  base  is  16.  What  is  the  diameter  of  the  base  of  an  equivalent 
cylinder  of  revolution,  whose  altitude  is  16  ? 

9.  The  area  of  the  entire  surface  of  a  frustum  of  a  cone  of  revo- 
lution is  306  Tc^  and  the  radii  of  its  bases  are  11  and  B.  Find  its  lat- 
eral area  and  volume. 

10 .  The  volume  of  a  frustum  of  a  cone  of  revolution  is  6020  n^  its 
altitude  is  60,  and  the  radius  of  its  lower  base  is  15.  Find  the  radius 
of  its  upper  base  and  its  lateral  area. 

11.  Find  the  altitude  and  lateral  area  of  a  cone  of  revolution, 
whose  volume  is  800  n,  and  whose  slant  height  is  to  the  diameter  of 
its  base  as  13  to  10. 


MEASUllEMENT  OF  THE  SPHERE. 


359 


THE  SPHERE. 


Definitions. 

679.  A  zone  is  a  portion  of  the  surface  of  a  sphere  in- 
cluded between  two  parallel  planes. 

The  circumferences  of  the  circles  which  bound  the  zone 
are  called  the  bases,  and  the  perpendicular  distance  between 
their  planes  the  altitude. 

A  zone  of  one  base  is  a  zone  one  of  whose  bounding  planes 
is  tangent  to  the  sphere. 

680.  A  spherical  segment  is  a  portion  of  a  sphere  included 
between  two  parallel  planes. 

The  circles  which  bound  it  are  called  the  bases,  and  the 
perpendicular  distance  between  them  the  altitude. 

A  spherical  segment  of  one  base  is  a  spherical  segment 
one  of  whose  boundiug  j^lanes  is  tangent  to  the  sphere. 

681.  If  the  semicircle  ACEB  be  revolved  about  its  diam- 
eter AB  as  an  axis,  and  CD  and  EF  are 
perpendicular  to  AB,  the  arc  CE  gener- 
ates a  zone  whose  altitude  i&'DF,  and 


the   figure   CEFD  a  spherical  segment 
whose  altitude  is  DF. 

The  arc  AC  generates  a  zone  of  one 
base,  and  the  figure  ACD  a  spherical 
segment  of  one  base. 

682.  If  a  semicircle  be  revolved  about 
its  diameter  as  an  axis,  any  sector  gener- 
ates a  solid  called  a  spherical  sector. 

Thus,  if  the  semicircle  ACDB  be  re- 
volved about  AB  as  an  axis,  the  sector 
OCD  generates  a  spherical  sector. 

The  zone  generated  by  the  arc  CD  is 
called  tlie  base  of  the  spherical  sector. 


360 


SOLID   GEOMETRY.  —  BOOK   IX. 


Proposition  IX.     Theorem. 

683.  The  area  generated  hy  the  revolution  of  a  straight 
line  about  an  axis  in  its  plane,  not  parallel  to  itself,  is  equal 
to  its  projection  on  the  axis,  multiplied  hy  the  circumference 
of  a  circle,  ivhose  radius  is  the  perpendicular  erected  at  tlte 
middle  point  of  the  line  and  terminating  in  the  axis. 


Let  the  straight  line  AB  be  revolved  about  the  axis  FM 
in  its  plane,  not  parallel  to  itself. 

Let  CD  be  the  projection  of  AB  on  FM,  and  EF  the  per- 
pendicular erected  at  the  middle  point  of  AB,  terminating 
in  the  axis. 

To  prove  area  AB*  =p  CDx2 ttEF.  (§  368.) 

Draw  AG  perpendicular  to  BD,  and  EH  perpendicular 
to  CD. 

The  surface  generated  \>y  AB  is  the  lateral  surface  of  a 
frustum  of  a  cone  of  revolution,  whose  bases  are  genei-ated 
hy  AC  and  BD. 

Then,  area  AB  =  AB  X  2  ttEH.  (§  675.) 

But  the  triangles  ABG  and  EFH  are  similai*.       (§  261.) 
AB  ^  EF 
AG 


Whence, 


EH 
Therefore,  AB  X EH  =  AGxEF 

=  CDxEF. 
Then,  area  AB  =  CD  X  2  ttEF. 


(§  231.) 


The  expression  '*  area  AB  "  is  used  to  denote  the  area  generated  by  AB. 


MEASUREMENT   OF   THE   SPHERE.  361 


Proposition  X.     Theokem. 

684.  If  an  isosceles  triangle  he  revolued  about  an  axis  in 
its  plane,  not  j^arallel  to  its  base,  which  passes  through  its 
vertex  ivlthout  iritersecthig  its  surface,  the  volume  generated 
is  equal  to  the  area  generated  by  the  base,  multiplied  by  one- 
third  the  altitude. 


Let  the  isosceles  triangle  OAB  be  revolved  about  the  axis 
OF  in  its  plane,  not  parallel  to  AB ;  and  draw  the  altitude 
OC. 

To  prove      vol.  OAB*  =  area  ABx^OC. 

Draw  AD  and  BE  perpendicular  to  OF-,  and  produce  BA 
to  meet  OF  at  F. 

Now,  vol.  OBF  =  vol.  OBF  -f-  vol.  BFF 

=  i  TT^R^'  X  0^  +  i  yrBF""  X  FF  (§  672.) 

=  ^  TrBF""  X  (OF  +  FF)  =  ^  irBF  X  BE  X  OF. 
But  BE  X  OF  =  OC  X  BF-,  for  each  expresses  twice  the 
area  of  the  triangle  OBF.  (§  313.) 

Hence,        vol.  OBF  =  ^  -kBE  xOC  xBF. 
But  irBE  X  BF  is  the  area  generated  by  BF.         (§  670.) 
Whence,        vol.  OBF  =  area  BF  X  ^  OC.  (1) 

Similarly,      voL  OAF  =  area  AF  X  ^  OC.  (2) 

Subtracting  (2)  from  (1),  we  have 

vol.  OAB  =  (area  BF  -  area  AF)  X  ^  OC 
=  area  AB  X  i  OC. 

♦  The  expression  *'  vol.  OAB  "  is  used  to  denote  the  volume  generated  by  OAB. 


362 


SOLID   GEOMETPvY.  —  BOOK  IX. 


Proposition  XI.     Theorem. 

685.    The  area  of  the  surface  of  a  sphere  is  equal  to  its 
diameter  multiplied  by  the  circumference  of  a  great  circle. 


I 


C/^ 

^ 

I  ^ 

E\^ 

.^ 

""^■^^^iii. 

Let  0  be  the  centre  of  the  semicircle  ADB ;  and  let  the 
sphere  be  generated  by  the  revolution  of  ADB  about  AB  as 
an  axis. 

Let  R  denote  the  radius  of  the  sphere. 

To  prove  that  the  area  of  the  surface  of  the  sphere  is 
AB  X^ttB.    . 

Divide  the  arc  ADB  into  four  equal  arcs,  AC,  CD,  DE, 
and  EB)  and  draw  the  chords  AC,  CD,  DE,  and  EB. 

Also,  draw  CC,  DO,  and  EE'  i)erpendicular  to  AB,  and 
OF  perpendicular  to  AC. 

Then,  area  AC  =  AC  X'^ -ttOF.  (§  683.) 

Also,  area  CD  =  CO  X  2tOF;  etc. 

Adding  these  equations,  we  have 
■   area  generated  by  broken  line  ACDEB 

=  (AC  +  C'O  +  etc.)  X  2  ttOF  =  AB  x  2  ttOF. 

Now  let  the  subdivisions  of  the  arc  ADB  be  bisected 
indefinitely. 

Then  the  area  generated -by  the  broken  line  ACDEB 
approaches  the  area  generated  by  the  arc  ADB  as  a  limit. 

(§  363,  L) 

And  AB  X  2  -nOF  approaches  AB  X  2  tt^  as  a  limit. 

•   (§  364,  1.) 

Then  area  generated  by  arc  ADB  =  AB  X  2  ttB.  (§  188.) 


MEASUREMENT  OF  THE   SPHERE.  363 

686.  Coii.  I.  Let  S  denote  the  area  of  tlie  surface  of  a 
sphere,  M  its  radius,  and  D  its  diameter. 

Then,  S  =  2Ex2  7rE  =  4.  ttB^ 

That  is,  the  area  of  the  surface  of  a  sphere  is  equal  to  the 
square  of  its  radius  inidtiplied  by  4  ir. 

Again,  S  =  ir  X  {"^  W  =  itD\ 

That  is,  the  area  of  the  surface  of  a  sphere  is  equal  to  the 
square  of  its  diameter  TnultipUed  hy  tt. 

687.  Cor.  II.  Let  S  and  8'  denote  the  areas  of  the 
surfaces  oi*  two  spheres,  i^  and  B!  their  radii,  and  B  and  1/ 
their  diameters. 

Then,  -1  ^  4^^  ^  ^ 

That  is,  the  areas  of  the  surfaces  of  two  spheres  are  to 
each  other  as  the  squares  of  their  radii,  or  as  the  squares  of 
their  diameters. 

688.  CoR.  III.  Let  0  be  the  centre  of  the  arc  AB ;  and 
draw  AA!  and  BB'  perpendicular  to  the  -^ 
diameter  OM.                                  ^                             A ^' 

It  may  be  proved,  as  in  §  685,  that  tlie  / 

area  generated  by  the   revolution  of  the        / 

arc  AB  about  OM  as  an  axis  is  equal  to  n 

AB'  x27r72, 
where  R  is  the  radius  of  the  arc. 

That  is,  the  area  of  a  zone  is  equal  to  its  altitude  viultiplied 
hy  the  circumference  of  a  great  circle. 

EXERCISES. 

12.  Find  the  area  of  the  surface  of  a  sphere  whose  radius  is  12. 

13.  Find  the  area  of  a  zone  whose  altitude  is  13,  if  the  radius  of 
the  sphere  is  1(5. 

14.  The  surface  of  a  sphere  is  equivalent  to  four  great  circles. 


364 


SOLID   GEOMETRY. —BOOK   IX. 


Propositiox   XII.     Theorem. 

689.    The  volume  of  a  sphere  is  equal  to  the  area  of  its 
surface  multiplied  by  one-third  its  radius. 


B   . 


Let  0  be  the  centre  of  the  semicircle  ADB ;  and  let  the 
sphere  be  generated  by  the  revolution  of  ADB  about  AB  as 
an  axis. 

Let  E  denote  the  radius  of  the  sphere. 

To  prove  that  the  volume  of  the  sphere  is  equal  to  the 
area  of  its  surface,  multiplied  by  ^  R. 

Divide  the  arc  ADB  into  four  equal  arcs,  AC,  CD,  DE, 
and  EB;  and  draw  the  chords  AC,  CD,  DE,  and  EB. 

Draw  OC,  OD,  and  OE-,  also,  OE  perpendicular  to  AC. 

Then,  vol.  OAC  =  area  AC  x  h  OE.  (§  684.) 

Also,  vol.  OCD  =  area  CDx  ^OE;  etc. 

Adding  these  equations,  we  have 

volume  generated  by  polygon  AC  DEB 

=  (area  AC  -\-  area  CD  +  etc.)  X  I  OE 
=  iiveiiACDEB  x  h  OE. 

Now  let  the  subdivisions  of  the  arc  ADB  be  bisected 
indefinitely. 

Then  the  volume  generated  by  the  polygon  AC  DEB 
approaches  the  volume  generated  by  the  semicircle  ADB 
as  a  limit.  (§  363,  II.) 

And  the  area  generated  by  A  CDEB  X  h  OE  approaches 
the  area  generated  by  the  arc  ADB  X  i  ^  as  a  limit. 

(§§  363,  I.,  364,  1.) 


MEASUREMENT   OF   THE   SPHERE.  365 

Then  volume  generated  by  ADB 

=  area  generated  by  arc  ADB  X  i  Ji.       (§  188.) 

690.  CoK.  I.  Let  V  denote  the  volume  of  a  sphere,  H 
its  radius,  and  D  its  diameter. 

Then,         V  =  4.  ttE'' X  ^  E^(%  686)  =  ^  ttR^ 
That  is,  the  volume  of  a  sphere  is  equal  to  the  cube  of  its 
radius  multiplied  by  |  tt. 

Again,  r  =J  tt  X  (2  ^)»  =  ^  TrZ)^. 

That  is,  the  volume  of  a  sphere  is  equal  to  the  cube  of  its 
diameter  multiplied  by  ^tt. 

691.  Cor.  II.  Let  V  and  V  denote  the  volumes  of  two 
spheres,  R  and  R'  their  radii,  and  D  and  //  their  diameters. 

Then  L-t£^  =  ^ 

That  is,  tlie  volumes  of  two  spheres  are  to  each  other  as  the 
cubes  of  their  radii,  or  as  the  cubes  of  tfieir  d'mmeters. 

692.  Cor.  III.    Let  OAB  be  a  sector  (jf  a  circle. 
It  may  be  proved,  as  in  §  689,  that  the 

volume  generated  by  the  revolution  of 
OAB  about  the  diameter  OM  as  an  axis, 
is  equal  to  the  area  generated  by  the  arc 
AB,  multiplied  by  ^  R,  where  R  is  the 
radius  of  the  arc. 

That  is,  tJie  volume  of  a  spherical  sector  Is  equal  to  the 
area  of  the  zone  which  forms  its  base,  inultiplied  by  one-third 
the  radius  of  the  sphere. 

693.  CoR.  IV.  Let  h  denote  the  altitude  of  the  zone 
which  forms  the  base  of  the  spherical  sector  of  §  692. 

Then,       vol.  OAB  =  h  X  2  irR  X  h  ^  (§  688.) 

=  I  ttRVi. 


366 


SOLID   GEOMETRY. —  BOOK   IX. 


694.  Cor.  V.  Let  P  denote  the  volume  of  a  spherical 
pyramid,  and  K  the  area  of  its  base ;  also,  let  V  denote  the 
volume,  S  the  area  of  the  surface,  and  B  the  radius,  of  the 
sphere. 


Then, 


^  (§  657) 


^ttR' 


(§§  686,090)  =\B. 


Whence,  P  =  K  X  I  R. 

That  is,  tJie  volume  of  a  spherical  pyramid  is  equal  to  the 
area  of  its  base  Tnultiplied  by  one-third  the  radius  of  the 
sjyhere. 


Pkoposition  XIII.     Problem. 
695.    To  find  the  volume  of  a  sj^herical  segment. 


A            1'' 

M 

»/fi\ 

0 

Let  0  be  the  centre  of  the  arc  ADB. 

Draw  AA'  and  BB'  perpendicular  to  the  diameter  OM. 

To  find  tlie  volume  of  the  spherical  segment  generated  by 
the  revolution  of  the  figure  ADBB'A'  about  031  as  an  axis. 

Let  AA'  =  /,  BB'  =  r,  A'B'  =  h,  and  OA  =  R. 

Draw  OA,  OB,  and  AB  ]  also,  draw  OC  perpendicular  to 
AB,  and  AE  perpendicular  to  BB\ 

:Now,  vol.  ADBB'A'  =  vol.  ACBD  +  vol.  ABB' A'.      (1 ) 


Also, 

But, 

And 


vol.  ACBD 


vol.  OADB 
irRVi. 
vol.  GAB  =  area  AB  X  h  OC 

X27rOC  XlOC 


vol.  OAB. 


vol.  OADB  =  I 


h 


(§  693.) 
(§  684.) 
(§  683.) 


I 


=  ^-^00 


MEASUREMENT   OF   THE   SPHERE.  367 

Then,       vol.  ACDB  =  |  ■jtR'^Ii  —  |  irOC^'h 

=  jf  TT  {R'  -  OC'')  h. 

But,  R''  -  OC'  =  AC''  _  (§  274.) 

Then,       vol.  ACDB  =  §0<  i  A^^'  X  /<^  =  ^  7r^Z?Vi. 
Now,  AB'  =  be""  +  AE"-  (§  273.) 

=  (r  _  /)2  4_  A2. 
Then,       vol.  ^(7Z>7?  =  ^  tt  [(r  -  /)2  +  A^]  /j. 
Also,       vol.  ABB' A'  =  J  tt  (/-^  +  /^  +  rr')  h.  (§  678.) 

Substituting  in  ( 1 ),  we  have 
vol.  ADBB'A' 

=  ^  ^  [(r  -  ry  +  A2]  /,  +  J  ^  (r2  +  /^  _^  ,./)  ^ 

=  ^  ^  (7-2—2  r?-'  +  7''2  4-  A2  _^  2  r'  +  2  /^  _^  2  r/)  /i 

=  ^  TT  (3  r2  +  3  /-^  +  A2)  /i 

=  i7r(r2  +  /2)A  +  ^^7,3. 

696.  CoR.  If  ?•  denotes  the  radius  of  the  base,  and  h  the 
altitude,  of  a  spherical  segment  of  one  base,  its  volume  is 

EXERCISES. 

15.  Find  the  volume  of  a  sphere  whose  radius  is  12. 

16.  Find  the  volume  of  a  spherical  sector  the  altitude  of  whose 
base  is  12,  the  diameter  of  the  sphere  being  25. 

17.  Find  the  volume  of  a  spherical  segment,  the  radii  of  whose 
bases  are  4  and  5,  and  whose  altitude  is  9. 

18.  Find  the  radius  and  volume  of  a  sphere  the  area  of  whose 
surface  is  324/1. 

19.  Find  the  diameter  and  area  of  the  surface  of  a  sphere  whose 
volume  is  ^^n. 

20.  The  sm-face  of  a  sphere  is  equivalent  to  the  lateral  surface  of 
its  circumscribed  cylinder. 

21.  The  volume  of  a  sphere  is  two-thirds  the  volume  of  its  cir- 
cumscribed cylinder. 

22.  A  spherical  cannon-ball  9  in.  in  diameter  is  dropped  into  a 
cubical  box  filled  with  water,  whose  depth  is  9  in.  How  many  cubic 
inches  of  water  will  be  left  in  the  box  ? 


368  SOLID   GEOMETRY.  —  BOOK  IX. 

23.  What  is  the  angle  of  the  base  of  a  spherical  wedge  whose 
volume  is  4j^  jt,  if  the  radius  of  the  sphere  is  4  ? 

24.  Find  the  area  of  a  spherical  triangle  whose  angles  are  125°, 
133°,  and  156°,  on  a  sphere  whose  radius  is  10. 

25.  Find  the  volume  of  a  quadrangular  spherical  pyramid,  the 
angles  of  whose  base  are  107°,  118°,  134°,  and  146°;  the  diameter  of 
the  sphere  being  12. 

26.  The  surface  of  a  sphere  is  equivalent  to  two-thirds  the  entire 
surface  of  its  circumscribed  cylinder. 

27.  The  volume  of  a  cylinder  of  revolution  is  equal  to  its  lateral 
area  multiplied  by  one-half  the  radius  of  its  base. 

28.  Prove  Prop.  IX.  when  the  straight  line  is  parallel  to  the  axis. 

29.  Find  the  area  of  the  surface  and  the  volume  of  a  sphere, 
inscribed  in  a  cube  the  area  of  whose  surface  is  486. 

30.  How  many  spherical  bullets,  each  ^  in.  in  diameter,  can  be 
formed  from  five  pieces  of  lead,  each  in  the  form  of  a  cone  of  revo- 
lution, the  radius  of  whose  base  is  5  in.,  and  whose  altitude  is  8  in.  ? 

31.  Find  the  volume  of  a  sphere  circumscribing  a  cube  whose 
volume  is  64. 

32.  A  cylindrical  vessel,  8  in.  in  diameter,  is  filled  to  the  brim 
with  water.  A  ball  is  immersed  in  it,  displacing  water  to  the  depth 
of  2^  in.     Find  the  diameter  of  the  ball. 

33.  The  radii  of  the  bases  of  two  similar  cylinders  of  revolution 
are  24  and  44,  respectively.  If  the  lateral  area  of  the  first  cylinder 
is  720,  what  is  the  lateral  area  of  the  second  ? 

34.  The  slant  heights  of  two  similar  cones  of  revolution  are  9 
and  15,  respectively.  If  the  volume  of  the  second  cone  is  625,  what 
is  the  volume  of  the  first  ? 

35.  The  total  areas  of  two  similar  cylinders  of  revolution  are  32 
and  162,  respectively.  If  the  volume  of  the  second  cylinder  is  1458, 
what  is  the  volume  of  the  first  ? 

36.  The  volumes  of  two  similar  cones  of  revolution  are  343  and 
512,  respectively.  If  the  lateral  area  of  the  first  cone  is  196,  what  is 
the  iateral  area  of  the  second  ? 

37.  If  a  sphere  6  in.  in  diameter  weighs  351  ounces,  what  is  the 
weight  of  a  sphere  of  the  same  material  whose  diameter  is  10  in.  ? 

38.  If  a  sphere  whose  radius  is  12^  in.  weighs  3125  lb.,  what  is 
the  radius  of  a  sphere  of  the  same  material  whose  weight  is  819^  lb.  ? 


MEASUREMENT   OF   THE   SPHERE.  869 

39.  The  altitude  of  a  frustum  of  a  cone  of  revolution  is  3|,  and 
the  radii  of  its  bases  are  5  and  3.  What  is  the  diameter  of  an 
equivalent  sphere  ? 

40.  Find  the  radius  of  a  sphere  whose  surface  is  equivalent  to  the 
entire  surface  of  a  cylinder  of  revolution,  whose  altitude  is  lOi  and 
radius  of  base  3. 

41.  A  cubical  piece  of  lead,  the  area  of  whose  entire  surface  is 
384  sq.  in.,  is  melted  and  formed  into  a  cone  of  revolution,  the  radius 
of  whose  base  is  12  in.     Find  the  altitude  of  the  cone. 

42.  The  volume  of  a  cylinder  of  revolution  is  equal  to  the  area  of 
its  generating  rectangle,  multiplied  by  the  circumference  of  a  circle 
whose  radius  is  the  distance  to  the  axis  from  the  centre  of  the 
rectangle. 

43.  The  volume  of  a  cone  of  revolution  is  equal  to  its  lateral 
area,  multiplied  by  one-third  the  distance  of  any  element  of  its 
lateral  surface  from  the  centre  of  the  base. 

44.  Two  zones  on  the  same  sphere,  or  on  equal  spheres,  are  to 
each  other  as  their  altitudes. 

45.  The  area  of  a  zone  of  one  base  is  equal  to  the  area  of  the 
circle  whose  radius  is  the  chord  of  its  generating  arc. 

46.  If  the  radius  of  a  sphere  is  i?,  what  is  the  area  of  a  zone  of 
one  base,  whose  generating  arc  is  45°  ? 

47.  If  the  altitude  of  a  cone  of  revolution  is  24,  and  its  slant 
height  25,  find  the  total  area  of  an  inscribed  cylinder,  the  radius  of 
whose  base  is  2. 

48.  Find  the  area  of  the  surface  and  the  volume  of  a  sphere  cir- 
cumscribing a  cylinder  of  revolution,  the  radius  of  whose  base  is  9, 
and  whose  altitude  is  24. 

49.  A  cone  of  revolution  is  circumscribed  about  a  sphere  whose 
diameter  is  two-thirds  the  altitude  of  the  cone.  Prove  that  its 
lateral  surface  and  volume  are,  respectively,  three-halves  and  nine- 
fourths  the  surface  and  volume  of  the  sphere. 

50.  If  the  altitude  of  a  cone  of  revolution  is  three-fourtlis  the 
radius  of  its  base,  its  volume  is  equal  to  its  lateral  area  multiplied  by 
one-fifth  the  radius  of  its  base. 

51.  A  cone  of  revolution  is  inscribed  in  a  sphere  whose  diameter 
is  f  the  altitude  of  the  cone.  Prove  that  its  lateral  surface  and  vol- 
ume are,  respectively,  I  and  /g  the  surface  and  volume  of  the  sphere. 


370  SOLID   GEOMETRY.  —  BOOK   IX. 


52.  If  the  radius  of  a  sphere  is  25,  find  the  lateral  area  and 
ume  of  an  inscribed  cone,  the  radius  of  whose  base  is  24. 

53.  If  the  volume  of  a  sphere  is  ^^  n,  find  the  lateral  area  and 
volume  of  a  circumscribed  cone  whose  altitude  is  18. 

54.  Find  the  volume  of  a  spherical  segment  of  one  base  whose 
altitude  is  6,  the  diameter  of  the  sphere  being  30. 

55.  A  circular  sector  whose  central  angle  is  45°  and  radius  12, 
revolves  about  a  diameter  perpendicular  to  one  of  its  bounding 
radii.     Find  the  volume  of  the  spherical  sector  generated. 

56.  Two  equal  small  circles  of  a  sphere  are  equally  distant  from 
the  centre. 

57.  A  square  whose  area  is  A,  revolves  about  its  diagonal  as  an 
axis.  Find  the  area  of  the  entire  surface,  and  the  volume,  of  the 
solid  generated. 

58.  How  many  cubic  feet  of  metal  are  there  in  a  hollow  cylindri- 
cal tube  18  ft.  long,  whose  outer  diameter  is  8  in.,  and  thickness 
1  in.  ? 

59.  The  altitude  of  a  cone  of  revolution  is  9  in.  At  what  dis- 
tances from  the  vertex  must  it  be  cut  by  planes  parallel  to  its  base, 
in  order  that  it  may  be  divided  into  three  equivalent  parts  ? 

60.  Given  the  radius  of  the  base,  R,  and  the  total  area,  T,  of  a 
cylinder  of  revolution,  to  find  its  volume. 

61.  Given  the  diameter  of  the  base,  D,  and  "the  volmne,  F,  of  a 
cylinder  of  revolution,  to  find  its  lateral  area  and  total  area. 

62.  Given  the  altitude,  //,  and  the  volume,  V,  of  a  cone  of  revo- 
lution, to  find  its  lateral  area. 

63.  Given  the  slant-height,  L,  and  the  lateral  area,  S,  of  a  cone 
of  revolution,  to  find  its  volume. 

64.  Given  the  area  of  the  surface  of  a  sphere,  S,  to  find  its 
volume. 

65.  Given  the  volume  of  a  sphere,  F,  to  find  the  area  of  its 
surface. 

66.  A  right  triangle  whose  legs  are  a  and  b  revolves  about  its 
hypotenuse  as  an  axis.  Find  the  area  of  the  entire  surface,  and  the 
volume,  of  the  solid  generated. 

67.  The  parallel  sides  of  a  trapezoid  are  12  and  33,  respectively, 
and  its  non-parallel  sides  are  10  and  17.  Find  the  volume  generated 
by  the  revolution  of  the  trapezoid  about  its  longest  side  as  an  axis^ 


vol-     ^ 


MEASUREMENT   OF    THE  SPHERE.  371 

68.  Find  the  diameter  of  a  sphere  in  which  the  area  of  the  sur- 
face and  the  volume  are  expressed  by  the  same  numbers. 

69.  An  equilateral  triangle,  whose  side  is  o,  revolves  about  one  of 
its  sides  as  an  axis.  Find  the  area  of  the  entire  surface,  and  the 
volume,  of  the  solid  generated. 

70.  An  equilateral  triangle,  whose  altitude  is  h,  revolves  about 
one  of  its  altitudes  as  an  axis.  Find  the  area  of  the  surface,  and  the 
volume,  of  the  solids  generated  by  the  triangle,  and  by  its  inscribed 
circle. 

71.  Find  the  lateral  area  and  volume  of  a  cylinder  of  revolution, 
whose  altitude  is  equal  to  the  diameter  of  its  base,  inscribed  in  a 
cone  of  revolution  whose  altitude  is  h,  and  radius  of  base  r. 

72.  Find  the  lateral  area  and  volimie  of  a  cylinder  of  revolution, 
whose  altitude  is  equal  to  the  diameter  of  its  base,  inscribed  in  a 
sphere  whose  radius  is  r. 

73.  An  equilateral  triangle,  whose  side  is  a,  revolves  about  a 
sti-aight  line  drawn  through  one  of  its  vertices  parallel  to  the  oppo- 
site side.  Find  the  area  of  the  entire  surface,  and  the  volume,  of 
the  solid  generated. 

74.  If  the  radius  of  a  sphere  is  7?,  find  the  circumference  and  area 
of  a  small  circle,  whose  distance  from  the  centre  is  h. 

75.  The  outer  diameter  of  a  spherical  shell  is  9  in.,  and  its  thick- 
ness is  1  in.     What  is  its  weight,  if  a  cubic  inch  of  the  metal  weighs 

76.  A  regular  hexagon,  whose  side  is  a,  revolves  about  its  longest 
diagonal  as  an  axis.  Find  the  area  of  the  entire  surface,  and  the 
volume,  of  the  solid  generated. 

77.  The  sides  AB  and  BC  of  a  rectangle  ABCD  are  5  and  8, 
respectively.  Find  the  volumes  generated  by  the  revolution  of  the 
triangle  ACD  about  the  sides  AB  and  BC  as  axes. 

78.  The  sides  of  a  triangle  are  17,  25,  and  28.  Find  the  volume 
generated  by  the  revolution  of  the  triangle  about  its  longest  side  as 
an  axis. 

79.  The  cross-section  of  a  tunnel  2^  miles  in  length  is  in 
the  form  of  a  rectangle  6  yd.  wide  and  4  yd.  high,  surmounted 
by  a  semicircle  whose  diameter  is  equal  to  the  width  of  the  rec- 
tangle. How  many  cubic  yards  of  material  were  taken  out  in  its 
construction  ? 


372  SOLID   GEOMETRY.— BOOK  IX. 

80.  A  frustum  of  a  circular  cone  is  equivalent  to  three  cones, 
whose  common  altitude  is  the  altitude  of  the  frustum,  and  wliose 
bases  are  the  lower  base,  the  upper  base,  and  a  mean  proportional 
between  the  bases  of  the  frustum.     (§  677.) 

81.  The  volume  of  a  cone  of  revolution  is  equal  to  the  area  of  its 
generating  triangle,  multiplied  by  the  circumference  of  a  circle 
whose  radius  is  the  distance  to  the  axis  from  the  intersection  of  the 
medians  of  the  triangle. 

82.  If  the  earth  be  regarded  as  a  sphere  whose  radius  is  B,  what 
is  the  area  of  the  zone  visible  from  a  point  whose  height  above  the 
surface  is  H  ? 

83.  The  sides  AB  and  BC  of  an  acute-angled  triangle  ABC,  are 
V241  and  10,  respectively.  Find  the  volume  generated  by  the  revo- 
lution of  the  triangle  about  an  axis  in  its  plane,  not  intersecting 
its  surface,  whose  distances  from  A,  B,  and  C  are  2,  17,  and  11, 
respectively. 

84.  A  projectile  consists  of  two  hemispheres,  connected  by  a  cyl- 
inder of  revolution.  If  the  altitude  and  diameter  of  the  base  of  the 
cylinder  are  8  in.  and  7  in.,  respectively,  find  the  number  of  cubic 
inches  in  the  projectile. 

85.  A  tapering  hollow  iron  column,  1  in.  thick,  is  24  ft.  long, 
10  in.  in  outside  diameter  at  one  end,  and  8  in.  in  diameter  at  the 
other.  How  many  cubic  inches  of  metal  were  used  in  its  construc- 
tion ? 

86.  If  any  triangle  be  revolved  about  an  axis  in  its  plane,  not 
parallel  to  its  base,  which  passes  through  its  vertex  without  inter- 
secting its  surface,  the  volume  generated  is  equal  to  the  area  gener- 
ated by  the  base,  multiplied  by  one-third  the  altitude. 

87.  If  any  triangle  be  revolved  about  an  axis  which  passes  through 
its  vertex  parallel  to  its  base,  the  volume  generated  is  equal  to  the 
area  generated  by  the  base,  multiplied  by  one-third  the  altitude. 

88.  A  segment  of  a  circle  whose  bounding  arc  is  a  quadrant,  and 
whose  radius  is  r,  revolves  about  a  diameter  parallel  to  its  bounding 
chord.  Find  the  area  of  the  entire  surface,  and  the  volume,  of  the 
solid  generated. 

89.  Find  the  area  of  the  surface  of  the  sphere  circumscribing  a 
regular  tetraedron  whose  edge  is  8. 


ANSWERS 


NUMERICAL    EXERCISES. 


Note.     Those  answers  are  omitted  which,  if  given,  would  destroy  the  utility 
of  the  problem. 


Page  16. 

6.   24°.         7.   63°^,  26°  30'.         8.   22°  SCT,  157°  30'. 

9.   37°. 

Page  41. 

27.  A  =  20°,  J?  =  60°,  C=  100°. 

Page  44. 

31.   ^  =  112°  30',  B  =  33°  45',  C  =  33°45'. 

Page  68. 

96.    7. 

Page  97. 

15.   52°  30'.        17.   96°.        18.    164°. 

Page  98. 

21.   28°  45'.  22.   44°  30'.  23.    12°. 

Page  99. 

25.   54°  30'.  26.    178°. 

Page  101. 

29.    112°  30'.  42.   83°,  89°  30',  97°,  90°  30',  74°  30'. 

Page  102. 

55.    Z  AEB  =  14°  30',    Z  AFB  =  10°  30'. 

58.    114°  30',  89°  30',  65°  30',  90°  30'. 

1 


2  GEOMETRY. 

Page  103. 

70.   97°  3(r,  89^  SCr,  82°  30',  9QP  3(r. 

Page  126. 

1.    112.        2.   42.         3.   f^.        4.   63. 

Pa«e  132. 

5.  Sh  2|.  6.    Ill,  18|. 

Page  138. 

7.    19|,  25^. 

Page  141. 
9.   4  ft.  6  in.  10.   12. 

Page  145. 

12.   15.  13.  37  ft.  1  in.  14.   47  ft.  6  in.  15.   4.33  +. 

16.   1  ft.  9.21  +  in. 

Page  147.     ^ 
18.   58.         19.   24. 

Page  153. 

21.   21.        24.    18.        27.   48.        28.   10.        29.   13^. 

30.    12.727  +.  31.   45. 

Page  154. 
33.    17f.        36.   50.        40.    Vl29,  2V2I,  V2OI.        41.   3^. 

Page  155. 

46.   36.        48.   63.        49.   3  and  4;   If  and  3^         55.   24. 

56.    17.        57,   21,  28.        58.   8V^.        59.   12,  4. 

Page  156. 

61.   ZVZ.  62.   14.  69.   21.  72.   70  and  99;  65  and  117. 

Page  165. 
1.   4:3. 

Page  167. 

2.   30|  ft.      3.   8  ft.  9  in.      4.    14,  12.       5.   6  ft.  11  in.,  20  ft.  9  inT 
6.   26  yd.  1  ft. 

Page  169. 
7.    6  sq.  ft.  60  sq.  in. 


ANSWERS. 


172. 
8.   2  sq.  ft.  48  sq.  in.  9.   243. 

Page  175. 
11.   210;   16^,  24H,  15.  12.   73.  13.    117. 


177. 

18.    \5.  ^,        19,   3V3.        20.    2  ft.  10  in.        22.    120.        25.    210. 

26.    18. 

Page  178. 

27.    l^ft.       28.   6.      29.   4v^.       30.    1260.       34.    120.       35.    l7. 

36.   624.        38.   540.         39.   28.         42.    4^.        43.   30,  10. 

Page  179. 

44.   36i.       47.   ^V2,  IIV2.       48.   .54.        51.   39,45.        52.    1010. 

53.   336. 

Page  207. 
29.  9.        30,   64:121.        31.    13.        32.   §V2. 

Page  211. 

34.    15.708,  19.635.         35.  Area,  452.-3^04. 

36.   Circumference,  50.2656.  49.   35.6048.  60.   35.81424. 

51.   9.827. 

Page  212. 
52.    10.2102.  53.   72.  54.    150.7968.  55.    1.2732. 

56.   201.0624.         57.    18.8496.         58.   50.2656. 

59.   37.6992,  9.4248.  60.    25.1328,  35.5377.  61.    9.06. 

62.    1306.9056  sq.  ft.         63.    120.99  ft.         64.   57  in. 

65.  57.295°+.        66.   2.658.        67.   5.64. 

Page  224. 
55.    IOV7. 

Page  225. 
61.    8.         65.    IH. 

Page  228. 
91.    480. 


4  •  GEOMETRY. 

Page  275. 
1.   4:3.         2.   2:5. 

Page  277. 

4.    42.  5.    1  ft.  9  in.  6.   34|i  cu.  in. ;  63|  sq.  in.  7.   574. 

8.    1008.        9.    12  and  7.         10.    1944.         13.    17. 

Page  279. 
14.   Volume,  50  VE. 

Page  280. 
16.   Volume,  ^V3. 


19.    v^,  18\^37,  180  V^.         20.   ^vTlS,  3Vl09,  15. 

21.    \/97,  I2V93,  72\/3.         22.    4\/39;  504  V3,  936  V3. 

23.    6V3,  56V26,  503^.  24.    4\/i0,  72V39,  672  VS. 

25.    150.  26.    320.  27.    2400  sq.  in.  28.   3if f  cu.  ft. 

29.   770.        30.   Volume,  48  VS.        31.   840.        32.   36  sq.  in. 

33.   12  in. 

Page  294. 

34.   512,  384.         35.    1705.         36,    144.        37.    10,  1. 

38.    700,  1568.         39.    iVEl,  640  V3.        40.   42A/9f,  624  Vs. 

41.   240,  ifaVIl9.        42.    108,  2lV39.        43.    768,2340. 

Page  295. 

49.   50.        52.   4V3,  IV2.         63.    15.         59.   438. 

Page  296. 

65.   9600  1b.         66.    168\/3,  15V219.         67.   5700  cu.  yd. 

68.   ^VSb. 

Page  308. 

73.    3456  cu.  in.        74.    6  ft.        75.   4  ft.  6  in.         76.   5v^in. 

77.   960,  3072.        78.    128.        79.    12.         80.    6.        84.   36V3. 

Page  337. 

11.   45°.         12.   4V3,  Vs. 

Page  344. 

13.   36.       14.   44.       15.   39f.      16.   86°  24'.       17.   3:2.       18.    108°. 

19.    220. 


I 


ANSWERS. 


346. 

22.   66^.        23.   364. 
Page  347. 

29.  60  cu.  ft.        30.    153°.        38.   30  in.,  8  in.,  20  in. 

Page  358. 

1.    28871,  4507r,  1296;r.     2.    175;r,  224n,  3927r.     3.  143/r,  2167r,  388/1 
4.    14,  12.         5.    28007r. 
6.    136  71.         7.   300  71.         8.   24.         9.    160  tt,  536  71. 
10.   4,  1159  71.         11.   24,  260  71. 

Page  363. 

12.   576  71.  13.   416  TT. 

Page  367. 

15.   2304  7r.  16.    1250  n.  17.   306  n.  18.    Volume,  972  n. 

19.   Area  of  surface,  225  n.  22.   347.2956  cu.  in. 

Page  368. 

23.   56°  15'.        24.   130  7f.         25.   58  7i.         29.   81  7r,  ^p  7r. 

30.  8192.         31.    32  n  VI.        32.    6  in.         33.   2420. 

34.    135.  35.    128.  36.  256.  37.    1625  oz.  38.   8  in. 

Page  369. 

39.    7.         40.    4^.         41.    -^  in.         46.    nR^  (2  -  V^). 
3/1 

47.   ^^  n.  48.    900  7z,  4500  n. 

Page  370. 

52.   960/1:,  6144  TT.         53.   ^ /r,  ^  ti.         54.   468/1. 

55.   576/1  \/2.  57.    nA  a/2,  \  nA  VIA.  58.   2.7489  +. 

59.   3  ^9  in.,  3  ^18  in.         60.    J^T-2nB^  ^ 


gj    4jr     8F+/tD'^  g2     V9F^  +  377g3^ 

'     D   '         2D        '  '  H 

g3_    l?V^^Y^^.  64.    S^.  65.    m-^. 

Zn'^L^  QVn 

66.   iL(±±^^,        ^«'^^"      .  67.    1216 /r. 

Va2  +  6--2       3  Va^  +  6^ 


GEOMETRY. 


Page  371. 

.69,    nofi  VS,  i  na^. 

70.    By  triangle,  nh^,  \  nh^\  by  inscribed  circle,  |  nli^,  -^^  nh^. 

'    (2  r  +  /O^  '  (2  r  +  /O^  ' 
'2.    2  7ir2,  1 7Tr3  \/2.  73^    2  Tia^  V3,  ^Tia^.  75.    67.3698  +  lb. 

76.   2  7ia2\/3,  7ia3.         77.    ^^n,^^n.        78.    2100  tt. 
79.   167803.68. 


82. 


2nBm 


li  +  II 
85.    7238.2464. 


Page  372. 
83.    1440  TT.  84.   487.4716. 

88.  2  7ir2  (1  +  >/2),  1 77r3  V2. 

89.  96  71. 


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